Focus Keyword: NCERT Solutions Class 9 Math Chapter 4 Exploring Algebraic Identities
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H1 Title: NCERT Solutions for Class 9 Math Chapter 4: Exploring Algebraic Identities (Exhaustive Step-by-Step Guide)
Navigating through the CBSE Class 9 Mathematics curriculum requires an in-depth understanding of algebraic expressions, polynomial expansions, geometric dissection models, and multi-variable factorisation techniques. Chapter 4 of Class 9 Mathematics, “Exploring Algebraic Identities”, forms the foundation of advanced coordinate geometry, quadratic equations, calculus, and mathematical modeling. It investigates the equality of algebraic expressions valid for all values of their variables; explores visual geometric proofs for binomial squares, trinomial expansions, and difference of squares; and details the methodical factorisation of complex cubic polynomials and conditional algebraic systems. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from foundational identities and geometric verifications to in-depth exercise sets, product evaluations, and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic mathematical workflows (Given Expression $\rightarrow$ Identity Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Factorised / Expanded Answer), and geometric inline SVG diagrams have been highlighted to ensure students secure maximum marks in their CBSE examinations.
Chapter 4: Exploring Algebraic Identities
Master Chapter Summary & Formula Blueprint
An algebraic identity is an algebraic equation that holds true for all possible values assigned to its variables. Unlike a conditional equation (which is satisfied only by specific roots), an identity represents an invariant equivalence used extensively for simplifying arithmetic calculations, expanding products, and factorising higher-degree polynomials.
| Identity Number | Algebraic Identity Formulation | Primary Operational Purpose | Typical Student Error | CBSE Marks Weightage |
|---|---|---|---|---|
| Identity I | $(x + y)^2 = x^2 + 2xy + y^2$ | Binomial sum expansion & mental math | Forgetting middle term $2xy$ as $x^2 + y^2$ | 1 to 2 Marks |
| Identity II | $(x – y)^2 = x^2 – 2xy + y^2$ | Binomial difference expansion | Writing negative sign on last term: $-y^2$ | 1 to 2 Marks |
| Identity III | $x^2 – y^2 = (x + y)(x – y)$ | Difference of two squares factorisation | Confusing $(x – y)^2$ with $x^2 – y^2$ | 2 to 3 Marks |
| Identity IV | $(x + a)(x + b) = x^2 + (a + b)x + ab$ | Product of binomials with common term | Sign errors when $a$ or $b$ is negative | 2 to 3 Marks |
| Identity V | $(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$ | Square of a trinomial | Dropping negative signs in cross-terms | 3 to 4 Marks |
| Identity VI | $(x + y)^3 = x^3 + y^3 + 3xy(x + y)$ | Expansion of cube of a sum | Omitting $3xy$ multiplier on $(x + y)$ | 3 to 4 Marks |
| Identity VII | $(x – y)^3 = x^3 – y^3 – 3xy(x – y)$ | Expansion of cube of a difference | Writing $+y^3$ or misapplying negative signs | 3 to 4 Marks |
| Identity VIII | $x^3 + y^3 + z^3 – 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ | Three-variable cubic factorisation | Incorrect sign sequence in second bracket | 4 to 5 Marks |
| Conditional Identity | If $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$ | Rapid arithmetic cubic evaluation | Applying without verifying $x + y + z = 0$ | 2 to 4 Marks |
🧠 Examiner’s Secret: When asked to factorise an expression like $4x^2 + 9y^2 + 16z^2 – 12xy – 24yz + 16zx$, look at the negative signs first. Here, the cross-terms involving $y$ (namely $-12xy$ and $-24yz$) are negative, while $16zx$ is positive. This implies that the term containing $y$ is negative, or both $x$ and $z$ are negative. The standard CBSE convention is to express this as $[2x + (-3y) + 4z]^2$.
Foundational Geometric Models & Identity Proofs
Geometric Area Model of Identity I: $(x + y)^2$
Identity I states that the square of the sum of two terms equals the sum of their squares plus twice their product: $(x + y)^2 = x^2 + 2xy + y^2$. x² xy xy y² x y x y
Consider a square whose side has length $(x + y)$.
- The total area of this large square is $(x + y) \times (x + y) = (x + y)^2$.
- By dividing the square with vertical and horizontal lines at distance $x$ from the top-left vertex, the large square is partitioned into four distinct sub-regions:
- A square with side $x$, having $\text{Area} = x^2$.
- A rectangle with dimensions $x \times y$, having $\text{Area} = xy$.
- A second rectangle with dimensions $y \times x$, having $\text{Area} = xy$.
- A small square with side $y$, having $\text{Area} = y^2$.
- Summing the areas of these four sub-regions gives:
$$\text{Total Area} = x^2 + xy + xy + y^2 = x^2 + 2xy + y^2$$ - Therefore:
$$(x + y)^2 = x^2 + 2xy + y^2$$
Geometric Dissection of Identity III: $x^2 – y^2$
Identity III states that the difference between the squares of two numbers equals the product of their sum and their difference: $x^2 – y^2 = (x + y)(x – y)$. Removed y² Area: x² – y² ➔ (x + y)(x – y)
Take a square of side $x$, whose total area is $x^2$. Cut out a smaller square of side $y$ from one of its corners, leaving an L-shaped region of area $x^2 – y^2$.
Dividing this remaining L-shaped region into two rectangular strips and rearranging them end-to-end creates a single large rectangle with length $(x + y)$ and breadth $(x – y)$.
Because the total area remains invariant under rearrangement:
$$x^2 – y^2 = (x + y)(x – y)$$
💡 Did You Know?: Ancient Babylonian mathematicians used Identity III over $3,500$ years ago to compute arithmetic multiplication quickly. Instead of directly calculating $a \times b$, they used the relation $ab = \frac{(a + b)^2 – (a – b)^2}{4}$, looking up the squares in pre-calculated numerical tables.
[👉 Also Read: Class 9 Math Chapter 2 Polynomials NCERT Solutions]
Step-by-Step Solutions: Core Textbook Exercises and Explorations
Question 1. Use suitable identities to find the following products:
(i) $(x + 4)(x + 10)$
(ii) $(x + 8)(x – 10)$
(iii) $(3x + 4)(3x – 5)$
(iv) $\left(y^2 + \frac{3}{2}\right)\left(y^2 – \frac{3}{2}\right)$
(v) $(3 – 2x)(3 + 2x)$
Answer:
(i) For $(x + 4)(x + 10)$:
- Identity Used: $(x + a)(x + b) = x^2 + (a + b)x + ab$
- Here, $a = 4$ and $b = 10$.
- Step-by-Step Substitution:
$$(x + 4)(x + 10) = x^2 + (4 + 10)x + (4 \times 10) = x^2 + 14x + 40$$ - Final Answer: $x^2 + 14x + 40$
(ii) For $(x + 8)(x – 10)$:
- Identity Used: $(x + a)(x + b) = x^2 + (a + b)x + ab$
- Here, $a = 8$ and $b = -10$.
- Step-by-Step Substitution:
$$(x + 8)(x – 10) = x^2 + [8 + (-10)]x + [8 \times (-10)] = x^2 – 2x – 80$$ - Final Answer: $x^2 – 2x – 80$
(iii) For $(3x + 4)(3x – 5)$:
- Identity Used: $(X + a)(X + b) = X^2 + (a + b)X + ab$, where $X = 3x$, $a = 4$, $b = -5$.
- Step-by-Step Substitution:
$$(3x + 4)(3x – 5) = (3x)^2 + 4 + (-5) + [4 \times (-5)]$$
$$= 9x^2 + (-1)(3x) – 20 = 9x^2 – 3x – 20$$ - Final Answer: $9x^2 – 3x – 20$
(iv) For $\left(y^2 + \frac{3}{2}\right)\left(y^2 – \frac{3}{2}\right)$:
- Identity Used: $(a + b)(a – b) = a^2 – b^2$
- Here, $a = y^2$ and $b = \frac{3}{2}$.
- Step-by-Step Substitution:
$$\left(y^2 + \frac{3}{2}\right)\left(y^2 – \frac{3}{2}\right) = (y^2)^2 – \left(\frac{3}{2}\right)^2 = y^4 – \frac{9}{4}$$ - Final Answer: $y^4 – \frac{9}{4}$
(v) For $(3 – 2x)(3 + 2x)$:
- Identity Used: $(a – b)(a + b) = a^2 – b^2$
- Here, $a = 3$ and $b = 2x$.
- Step-by-Step Substitution:
$$(3 – 2x)(3 + 2x) = 3^2 – (2x)^2 = 9 – 4x^2$$ - Final Answer: $9 – 4x^2$
Question 2. Evaluate the following products without multiplying directly:
(i) $103 \times 107$
(ii) $95 \times 96$
(iii) $104 \times 96$
Answer:
(i) For $103 \times 107$:
- Express the factors as binomials around $100$:
$$103 \times 107 = (100 + 3)(100 + 7)$$ - Identity Used: $(x + a)(x + b) = x^2 + (a + b)x + ab$
- Substitute $x = 100$, $a = 3$, $b = 7$:
$$(100 + 3)(100 + 7) = 100^2 + (3 + 7)(100) + (3 \times 7)$$
$$= 10000 + (10)(100) + 21 = 10000 + 1000 + 21 = 11021$$ - Final Answer: $11021$
(ii) For $95 \times 96$:
- Express the factors around $100$:
$$95 \times 96 = (100 – 5)(100 – 4)$$ - Identity Used: $(x + a)(x + b) = x^2 + (a + b)x + ab$
- Substitute $x = 100$, $a = -5$, $b = -4$:
$$(100 – 5)(100 – 4) = 100^2 + (-5) + (-4) + [(-5) \times (-4)]$$
$$= 10000 + (-9)(100) + 20 = 10000 – 900 + 20 = 9120$$ - Final Answer: $9120$
(iii) For $104 \times 96$:
- Express the factors around $100$:
$$104 \times 96 = (100 + 4)(100 – 4)$$ - Identity Used: $(a + b)(a – b) = a^2 – b^2$
- Substitute $a = 100$, $b = 4$:
$$(100 + 4)(100 – 4) = 100^2 – 4^2 = 10000 – 16 = 9984$$ - Final Answer: $9984$
Question 3. Factorise the following using appropriate identities:
(i) $9x^2 + 6xy + y^2$
(ii) $4y^2 – 4y + 1$
(iii) $x^2 – \frac{y^2}{100}$
Answer:
(i) For $9x^2 + 6xy + y^2$:
- Rewrite each term in squared form:
$$9x^2 = (3x)^2, \quad y^2 = (y)^2, \quad 6xy = 2(3x)(y)$$ - The expression matches the expansion $a^2 + 2ab + b^2 = (a + b)^2$.
- Here, $a = 3x$ and $b = y$:
$$(3x)^2 + 2(3x)(y) + (y)^2 = (3x + y)^2 = (3x + y)(3x + y)$$ - Final Answer: $(3x + y)(3x + y)$
(ii) For $4y^2 – 4y + 1$:
- Rewrite terms in squared form:
$$4y^2 = (2y)^2, \quad 1 = (1)^2, \quad 4y = 2(2y)(1)$$ - The expression matches $a^2 – 2ab + b^2 = (a – b)^2$.
- Here, $a = 2y$ and $b = 1$:
$$(2y)^2 – 2(2y)(1) + (1)^2 = (2y – 1)^2 = (2y – 1)(2y – 1)$$ - Final Answer: $(2y – 1)(2y – 1)$
(iii) For $x^2 – \frac{y^2}{100}$:
- Express both terms as perfect squares:
$$x^2 – \left(\frac{y}{10}\right)^2$$ - Identity Used: $a^2 – b^2 = (a + b)(a – b)$
- Substitute $a = x$ and $b = \frac{y}{10}$:
$$x^2 – \left(\frac{y}{10}\right)^2 = \left(x + \frac{y}{10}\right)\left(x – \frac{y}{10}\right)$$ - Final Answer: $\left(x + \frac{y}{10}\right)\left(x – \frac{y}{10}\right)$
Question 4. Expand each of the following using suitable identities:
(i) $(x + 2y + 4z)^2$
(ii) $(2x – y + z)^2$
(iii) $(-2x + 3y + 2z)^2$
(iv) $(3a – 7b – c)^2$
(v) $(-2x + 5y – 3z)^2$
(vi) $\left[\frac{1}{4}a – \frac{1}{2}b + 1\right]^2$
Answer:
General Identity (Square of a Trinomial):
$$(A + B + C)^2 = A^2 + B^2 + C^2 + 2AB + 2BC + 2CA$$
(i) For $(x + 2y + 4z)^2$:
- $A = x, B = 2y, C = 4z$
$$(x + 2y + 4z)^2 = x^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$$
$$= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$$ - Final Answer: $x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$
(ii) For $(2x – y + z)^2$:
- $A = 2x, B = -y, C = z$
$$(2x – y + z)^2 = (2x)^2 + (-y)^2 + z^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$$
$$= 4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx$$ - Final Answer: $4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx$
(iii) For $(-2x + 3y + 2z)^2$:
- $A = -2x, B = 3y, C = 2z$
$$(-2x + 3y + 2z)^2 = (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)$$
$$= 4x^2 + 9y^2 + 4z^2 – 12xy + 12yz – 8zx$$ - Final Answer: $4x^2 + 9y^2 + 4z^2 – 12xy + 12yz – 8zx$
(iv) For $(3a – 7b – c)^2$:
- $A = 3a, B = -7b, C = -c$
$$(3a – 7b – c)^2 = (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$$
$$= 9a^2 + 49b^2 + c^2 – 42ab + 14bc – 6ca$$ - Final Answer: $9a^2 + 49b^2 + c^2 – 42ab + 14bc – 6ca$
(v) For $(-2x + 5y – 3z)^2$:
- $A = -2x, B = 5y, C = -3z$
$$(-2x + 5y – 3z)^2 = (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)$$
$$= 4x^2 + 25y^2 + 9z^2 – 20xy – 30yz + 12zx$$ - Final Answer: $4x^2 + 25y^2 + 9z^2 – 20xy – 30yz + 12zx$
(vi) For $\left[\frac{1}{4}a – \frac{1}{2}b + 1\right]^2$:
- $A = \frac{1}{4}a, B = -\frac{1}{2}b, C = 1$
$$\left[\frac{1}{4}a – \frac{1}{2}b + 1\right]^2 = \left(\frac{1}{4}a\right)^2 + \left(-\frac{1}{2}b\right)^2 + 1^2 + 2\left(\frac{1}{4}a\right)\left(-\frac{1}{2}b\right) + 2\left(-\frac{1}{2}b\right)(1) + 2(1)\left(\frac{1}{4}a\right)$$
$$= \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 – \frac{1}{4}ab – b + \frac{1}{2}a$$ - Final Answer: $\frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 – \frac{1}{4}ab – b + \frac{1}{2}a$
Question 5. Factorise:
(i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
(ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
Answer:
(i) For $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$:
- Inspect the signs of the cross-terms:
The terms $-24yz$ and $-16xz$ are both negative, while $+12xy$ is positive.
This shows that the variable $z$ carries a negative sign:
$$4x^2 = (2x)^2, \quad 9y^2 = (3y)^2, \quad 16z^2 = (-4z)^2$$ - Verify cross-products:
$$2(2x)(3y) = 12xy$$
$$2(3y)(-4z) = -24yz$$
$$2(-4z)(2x) = -16xz$$ - This matches $(A + B + C)^2$:
$$[2x + 3y + (-4z)]^2 = (2x + 3y – 4z)^2$$ - Final Answer: $(2x + 3y – 4z)(2x + 3y – 4z)$
(ii) For $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$:
- Inspect signs of cross-terms:
The terms $-2\sqrt{2}xy$ and $-8xz$ are negative, while $+4\sqrt{2}yz$ is positive.
This shows that the term containing $x$ carries a negative sign. - Express terms as squares:
$$2x^2 = (-\sqrt{2}x)^2, \quad y^2 = (y)^2, \quad 8z^2 = (2\sqrt{2}z)^2$$ - Verify cross-products:
$$2(-\sqrt{2}x)(y) = -2\sqrt{2}xy$$
$$2(y)(2\sqrt{2}z) = 4\sqrt{2}yz$$
$$2(2\sqrt{2}z)(-\sqrt{2}x) = 2(-2 \times 2)xz = -8xz$$ - Matching with $(A + B + C)^2$:
$$(-\sqrt{2}x + y + 2\sqrt{2}z)^2$$ - Final Answer: $(-\sqrt{2}x + y + 2\sqrt{2}z)(-\sqrt{2}x + y + 2\sqrt{2}z)$ (or $(\sqrt{2}x – y – 2\sqrt{2}z)^2$)
Question 6. Write the following cubes in expanded form:
(i) $(2x + 1)^3$
(ii) $(2a – 3b)^3$
(iii) $\left[\frac{3}{2}x + 1\right]^3$
(iv) $\left[x – \frac{2}{3}y\right]^3$
Answer:
Identities for Cubes of Binomials:
$$(A + B)^3 = A^3 + B^3 + 3AB(A + B) = A^3 + 3A^2B + 3AB^2 + B^3$$
$$(A – B)^3 = A^3 – B^3 – 3AB(A – B) = A^3 – 3A^2B + 3AB^2 – B^3$$
(i) For $(2x + 1)^3$:
- $A = 2x, B = 1$
$$(2x + 1)^3 = (2x)^3 + 1^3 + 3(2x)(1)(2x + 1)$$
$$= 8x^3 + 1 + 6x(2x + 1) = 8x^3 + 1 + 12x^2 + 6x$$
$$= 8x^3 + 12x^2 + 6x + 1$$ - Final Answer: $8x^3 + 12x^2 + 6x + 1$
(ii) For $(2a – 3b)^3$:
- $A = 2a, B = 3b$
$$(2a – 3b)^3 = (2a)^3 – (3b)^3 – 3(2a)(3b)(2a – 3b)$$
$$= 8a^3 – 27b^3 – 18ab(2a – 3b) = 8a^3 – 27b^3 – 36a^2b + 54ab^2$$
$$= 8a^3 – 36a^2b + 54ab^2 – 27b^3$$ - Final Answer: $8a^3 – 36a^2b + 54ab^2 – 27b^3$
(iii) For $\left[\frac{3}{2}x + 1\right]^3$:
- $A = \frac{3}{2}x, B = 1$
$$\left[\frac{3}{2}x + 1\right]^3 = \left(\frac{3}{2}x\right)^3 + 1^3 + 3\left(\frac{3}{2}x\right)(1)\left(\frac{3}{2}x + 1\right)$$
$$= \frac{27}{8}x^3 + 1 + \frac{9}{2}x\left(\frac{3}{2}x + 1\right) = \frac{27}{8}x^3 + 1 + \frac{27}{4}x^2 + \frac{9}{2}x$$
$$= \frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$$ - Final Answer: $\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$
(iv) For $\left[x – \frac{2}{3}y\right]^3$:
- $A = x, B = \frac{2}{3}y$
$$\left[x – \frac{2}{3}y\right]^3 = x^3 – \left(\frac{2}{3}y\right)^3 – 3(x)\left(\frac{2}{3}y\right)\left(x – \frac{2}{3}y\right)$$
$$= x^3 – \frac{8}{27}y^3 – 2xy\left(x – \frac{2}{3}y\right) = x^3 – \frac{8}{27}y^3 – 2x^2y + \frac{4}{3}xy^2$$
$$= x^3 – 2x^2y + \frac{4}{3}xy^2 – \frac{8}{27}y^3$$ - Final Answer: $x^3 – 2x^2y + \frac{4}{3}xy^2 – \frac{8}{27}y^3$
Question 7. Evaluate the following using suitable identities:
(i) $99^3$
(ii) $102^3$
(iii) $998^3$
Answer:
(i) For $99^3$:
- Express as $(100 – 1)^3$:
$$(100 – 1)^3 = 100^3 – 1^3 – 3(100)(1)(100 – 1)$$
$$= 1000000 – 1 – 300(99) = 1000000 – 1 – 29700$$
$$= 1000000 – 29701 = 970299$$ - Final Answer: $970299$
(ii) For $102^3$:
- Express as $(100 + 2)^3$:
$$(100 + 2)^3 = 100^3 + 2^3 + 3(100)(2)(100 + 2)$$
$$= 1000000 + 8 + 600(102) = 1000000 + 8 + 61200 = 1061208$$ - Final Answer: $1061208$
(iii) For $998^3$:
- Express as $(1000 – 2)^3$:
$$(1000 – 2)^3 = 1000^3 – 2^3 – 3(1000)(2)(1000 – 2)$$
$$= 1000000000 – 8 – 6000(998)$$
$$= 1000000000 – 8 – 5988000 = 1000000000 – 5988008 = 994011992$$ - Final Answer: $994011992$
Question 8. Factorise each of the following:
(i) $8a^3 + b^3 + 12a^2b + 6ab^2$
(ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
(iii) $27 – 125a^3 – 135a + 225a^2$
(iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
(v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
Answer:
(i) For $8a^3 + b^3 + 12a^2b + 6ab^2$:
- Rewrite the cubed terms:
$$8a^3 = (2a)^3, \quad b^3 = (b)^3$$ - Group the remaining terms:
$$12a^2b + 6ab^2 = 3(2a)(b)(2a + b)$$ - This matches $A^3 + B^3 + 3AB(A + B) = (A + B)^3$ with $A = 2a, B = b$.
$$(2a + b)^3 = (2a + b)(2a + b)(2a + b)$$ - Final Answer: $(2a + b)(2a + b)(2a + b)$
(ii) For $8a^3 – b^3 – 12a^2b + 6ab^2$:
- Rewrite terms:
$$(2a)^3 – (b)^3 – 3(2a)(b)(2a – b) = (2a – b)^3$$ - Final Answer: $(2a – b)(2a – b)(2a – b)$
(iii) For $27 – 125a^3 – 135a + 225a^2$:
- Rewrite terms:
$$27 = 3^3, \quad 125a^3 = (5a)^3$$
$$-135a + 225a^2 = -3(3)(5a)(3 – 5a)$$ - This matches $A^3 – B^3 – 3AB(A – B) = (A – B)^3$ with $A = 3, B = 5a$:
$$(3 – 5a)^3 = (3 – 5a)(3 – 5a)(3 – 5a)$$ - Final Answer: $(3 – 5a)(3 – 5a)(3 – 5a)$
(iv) For $64a^3 – 27b^3 – 144a^2b + 108ab^2$:
- Rewrite terms:
$$64a^3 = (4a)^3, \quad 27b^3 = (3b)^3$$
$$-144a^2b + 108ab^2 = -3(4a)(3b)(4a – 3b)$$ - This matches $(A – B)^3$ with $A = 4a, B = 3b$:
$$(4a – 3b)^3 = (4a – 3b)(4a – 3b)(4a – 3b)$$ - Final Answer: $(4a – 3b)(4a – 3b)(4a – 3b)$
(v) For $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$:
- Rewrite terms:
$$27p^3 = (3p)^3, \quad \frac{1}{216} = \left(\frac{1}{6}\right)^3$$
$$-\frac{9}{2}p^2 + \frac{1}{4}p = -3(3p)\left(\frac{1}{6}\right)\left(3p – \frac{1}{6}\right)$$ - This matches $(A – B)^3$ with $A = 3p, B = \frac{1}{6}$:
$$\left(3p – \frac{1}{6}\right)^3 = \left(3p – \frac{1}{6}\right)\left(3p – \frac{1}{6}\right)\left(3p – \frac{1}{6}\right)$$ - Final Answer: $\left(3p – \frac{1}{6}\right)\left(3p – \frac{1}{6}\right)\left(3p – \frac{1}{6}\right)$
Question 9. Verify:
(i) $x^3 + y^3 = (x + y)(x^2 – xy + y^2)$
(ii) $x^3 – y^3 = (x – y)(x^2 + xy + y^2)$
Answer:
(i) Verification of $x^3 + y^3 = (x + y)(x^2 – xy + y^2)$:
- Start with the Right-Hand Side ($\text{RHS}$):
$$\text{RHS} = (x + y)(x^2 – xy + y^2)$$ - Expand using the distributive property:
$$\text{RHS} = x(x^2 – xy + y^2) + y(x^2 – xy + y^2)$$
$$= (x^3 – x^2y + xy^2) + (yx^2 – xy^2 + y^3)$$ - Combine like terms:
$$\text{RHS} = x^3 + (-x^2y + x^2y) + (xy^2 – xy^2) + y^3$$
$$\text{RHS} = x^3 + 0 + 0 + y^3 = x^3 + y^3 = \text{LHS}$$ - Conclusion: Hence Verified.
(ii) Verification of $x^3 – y^3 = (x – y)(x^2 + xy + y^2)$:
- Start with the Right-Hand Side ($\text{RHS}$):
$$\text{RHS} = (x – y)(x^2 + xy + y^2)$$ - Expand using the distributive property:
$$\text{RHS} = x(x^2 + xy + y^2) – y(x^2 + xy + y^2)$$
$$= (x^3 + x^2y + xy^2) – (x^2y + xy^2 + y^3)$$
$$= x^3 + x^2y + xy^2 – x^2y – xy^2 – y^3$$ - Combine like terms:
$$\text{RHS} = x^3 + (x^2y – x^2y) + (xy^2 – xy^2) – y^3$$
$$\text{RHS} = x^3 + 0 + 0 – y^3 = x^3 – y^3 = \text{LHS}$$ - Conclusion: Hence Verified.
Question 10. Factorise each of the following:
(i) $27y^3 + 125z^3$
(ii) $64m^3 – 343n^3$
Answer:
(i) For $27y^3 + 125z^3$:
- Express each term as a cube:
$$27y^3 = (3y)^3, \quad 125z^3 = (5z)^3$$ - Identity Used: $a^3 + b^3 = (a + b)(a^2 – ab + b^2)$
- Substitute $a = 3y, b = 5z$:
$$(3y)^3 + (5z)^3 = (3y + 5z)[(3y)^2 – (3y)(5z) + (5z)^2]$$
$$= (3y + 5z)(9y^2 – 15yz + 25z^2)$$ - Final Answer: $(3y + 5z)(9y^2 – 15yz + 25z^2)$
(ii) For $64m^3 – 343n^3$:
- Express each term as a cube:
$$64m^3 = (4m)^3, \quad 343n^3 = (7n)^3$$ - Identity Used: $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$
- Substitute $a = 4m, b = 7n$:
$$(4m)^3 – (7n)^3 = (4m – 7n)[(4m)^2 + (4m)(7n) + (7n)^2]$$
$$= (4m – 7n)(16m^2 + 28mn + 49n^2)$$ - Final Answer: $(4m – 7n)(16m^2 + 28mn + 49n^2)$
Question 11. Factorise: $27x^3 + y^3 + z^3 – 9xyz$
Answer:
Step 1: Identify Relevant Identity
The expression matches the left-hand side of Identity VIII:
$$a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)$$
Step 2: Rewrite in Cubed Form
- $27x^3 = (3x)^3$
- $y^3 = (y)^3$
- $z^3 = (z)^3$
- $-9xyz = -3(3x)(y)(z)$
Step 3: Step-by-Step Substitution
Here, $a = 3x, b = y, c = z$:
$$(3x)^3 + y^3 + z^3 – 3(3x)(y)(z)$$
$$= (3x + y + z)[(3x)^2 + y^2 + z^2 – (3x)(y) – (y)(z) – (z)(3x)]$$
$$= (3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)$$
Final Answer:
$(3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)$
Question 12. Verify that:
$$x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$$
Answer:
Step 1: Start with the Right-Hand Side ($\text{RHS}$)
$$\text{RHS} = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$$
Step 2: Expand the Squared Terms Inside the Bracket
$$(x – y)^2 = x^2 – 2xy + y^2$$
$$(y – z)^2 = y^2 – 2yz + z^2$$
$$(z – x)^2 = z^2 – 2zx + x^2$$
Step 3: Combine Expanded Terms
$$(x – y)^2 + (y – z)^2 + (z – x)^2 = (x^2 + x^2) + (y^2 + y^2) + (z^2 + z^2) – 2xy – 2yz – 2zx$$
$$= 2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx$$
$$= 2(x^2 + y^2 + z^2 – xy – yz – zx)$$
Step 4: Multiply by $\frac{1}{2}(x + y + z)$
$$\text{RHS} = \frac{1}{2}(x + y + z) \times 2(x^2 + y^2 + z^2 – xy – yz – zx)$$
$$= (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)$$
By Identity VIII, this product equals:
$$x^3 + y^3 + z^3 – 3xyz = \text{LHS}$$
Conclusion:
Hence Verified.
Question 13. If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
Answer:
Step 1: State Identity VIII
$$x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)$$
Step 2: Substitute Given Condition $x + y + z = 0$
$$x^3 + y^3 + z^3 – 3xyz = (0) \times (x^2 + y^2 + z^2 – xy – yz – zx)$$
$$x^3 + y^3 + z^3 – 3xyz = 0$$
Step 3: Rearrange to Final Form
Add $3xyz$ to both sides:
$$x^3 + y^3 + z^3 = 3xyz$$
Conclusion:
Hence Proved.
Question 14. Without actually calculating the cubes, find the value of each of the following:
(i) $(-12)^3 + 7^3 + 5^3$
(ii) $28^3 + (-15)^3 + (-13)^3$
Answer:
(i) For $(-12)^3 + 7^3 + 5^3$:
- Let $x = -12, y = 7, z = 5$.
- Test the conditional sum $x + y + z$:
$$x + y + z = -12 + 7 + 5 = -12 + 12 = 0$$ - Since $x + y + z = 0$, by the conditional identity:
$$x^3 + y^3 + z^3 = 3xyz$$ - Step-by-Step Calculation:
$$(-12)^3 + 7^3 + 5^3 = 3(-12)(7)(5) = 3 \times (-12) \times 35 = -36 \times 35 = -1260$$ - Final Answer: $-1260$
(ii) For $28^3 + (-15)^3 + (-13)^3$:
- Let $x = 28, y = -15, z = -13$.
- Test the conditional sum $x + y + z$:
$$x + y + z = 28 + (-15) + (-13) = 28 – 28 = 0$$ - Since $x + y + z = 0$:
$$x^3 + y^3 + z^3 = 3xyz$$ - Step-by-Step Calculation:
$$28^3 + (-15)^3 + (-13)^3 = 3(28)(-15)(-13) = 84 \times 195 = 16380$$ - Final Answer: $16380$
Question 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) $\text{Area} = 25a^2 – 35a + 12$
(ii) $\text{Area} = 35y^2 + 13y – 12$
Answer:
(i) For $\text{Area} = 25a^2 – 35a + 12$:
- Factorise by splitting the middle term.
Product $p \times q = 25 \times 12 = 300$, Sum $p + q = -35$.
The two numbers are $-20$ and $-15$ (since $(-20) \times (-15) = 300$ and $(-20) + (-15) = -35$). - Split the middle term:
$$25a^2 – 35a + 12 = 25a^2 – 20a – 15a + 12$$
$$= 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)$$ - Since $\text{Area} = \text{Length} \times \text{Breadth}$:
One factor represents the length, and the other represents the breadth. - Final Answer:
$\text{Length} = (5a – 3), \quad \text{Breadth} = (5a – 4)$ (or vice versa)
(ii) For $\text{Area} = 35y^2 + 13y – 12$:
- Factorise by splitting the middle term.
Product $p \times q = 35 \times (-12) = -420$, Sum $p + q = 13$.
The two numbers are $28$ and $-15$ (since $28 \times (-15) = -420$ and $28 – 15 = 13$). - Split the middle term:
$$35y^2 + 13y – 12 = 35y^2 + 28y – 15y – 12$$
$$= 7y(5y + 4) – 3(5y + 4) = (5y + 4)(7y – 3)$$ - Final Answer:
$\text{Length} = (5y + 4), \quad \text{Breadth} = (7y – 3)$ (or vice versa)
Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) $\text{Volume} = 3x^2 – 12x$
(ii) $\text{Volume} = 12ky^2 + 8ky – 20k$
Answer:
(i) For $\text{Volume} = 3x^2 – 12x$:
- Factor out common terms:
$$3x^2 – 12x = 3x(x – 4) = 3 \times x \times (x – 4)$$ - Since $\text{Volume of Cuboid} = \text{length} \times \text{breadth} \times \text{height}$:
- Final Answer:
Possible dimensions are $3$, $x$, and $(x – 4)$.
(ii) For $\text{Volume} = 12ky^2 + 8ky – 20k$:
- Factor out the greatest common constant factor $4k$:
$$12ky^2 + 8ky – 20k = 4k(3y^2 + 2y – 5)$$ - Factorise the quadratic polynomial $3y^2 + 2y – 5$ by splitting the middle term:
Product $= 3 \times (-5) = -15$, Sum $= 2$. Numbers are $5$ and $-3$:
$$3y^2 + 5y – 3y – 5 = y(3y + 5) – 1(3y + 5) = (3y + 5)(y – 1)$$ - Combine all factors:
$$\text{Volume} = 4k(3y + 5)(y – 1)$$ - Final Answer:
Possible dimensions are $4k$, $(3y + 5)$, and $(y – 1)$.
[👉 Also Read: Class 9 Math Chapter 5 Introduction to Euclid’s Geometry NCERT Solutions]
Master High-Yield Board FAQs (Rank Math Schema Ready)
What is the primary difference between an algebraic identity and an algebraic equation?
An algebraic identity is an equality that holds true for every possible numerical value assigned to its variables (e.g., $(x+1)^2 = x^2+2x+1$). In contrast, an algebraic equation is a conditional statement that holds true only for specific root values (e.g., $x^2 – 5x + 6 = 0$ is true only when $x = 2$ or $x = 3$).
How do you remember the sign placements in $(x – y)^3$?
The expansion of $(x – y)^3$ follows an alternating sign pattern: $x^3 – 3x^2y + 3xy^2 – y^3$. Any term containing an odd power of $y$ ($y^1$ and $y^3$) carries a negative sign, while terms with even powers ($y^0$ and $y^2$) carry a positive sign.
What is the conditional identity for the sum of three cubes?
The conditional identity states that if the sum of three terms is zero ($x + y + z = 0$), then the sum of their cubes equals three times their product: $x^3 + y^3 + z^3 = 3xyz$. This eliminates the need to compute large cubes directly.
How do you factorise $x^3 + y^3$ and $x^3 – y^3$?
The sum of two cubes factorises as $x^3 + y^3 = (x + y)(x^2 – xy + y^2)$, and the difference of two cubes factorises as $x^3 – y^3 = (x – y)(x^2 + xy + y^2)$. Note that the sign in the linear binomial matches the operation, while the cross-term in the quadratic factor takes the opposite sign.
What geometric figure represents Identity I: $(x + y)^2$?
Identity I is represented geometrically by a square of side length $(x + y)$. Dividing this square into four sections produces one square of area $x^2$, one square of area $y^2$, and two identical rectangles of area $xy$, demonstrating visually that $(x + y)^2 = x^2 + 2xy + y^2$.
Why does $(x + a)(x + b)$ equal $x^2 + (a + b)x + ab$?
Applying the distributive property yields $x(x + b) + a(x + b) = x^2 + bx + ax + ab$. Factoring out the common variable $x$ from the two middle terms gives $x^2 + (a + b)x + ab$.
How do you identify which variable carries the negative sign in an expanded trinomial square?
To find the negative variable in an expansion like $a^2 + b^2 + c^2 – 2ab + 2bc – 2ca$, identify the two negative cross-terms (here, $-2ab$ and $-2ca$). The variable common to both negative terms (in this case, $a$) is the one that carries the negative sign.
Can algebraic identities be used to multiply numbers with decimals?
Yes. Decimals can be expressed around a convenient base. For example, $10.2 \times 9.8$ can be written as $(10 + 0.2)(10 – 0.2)$. Applying Identity III ($a^2 – b^2$) yields $10^2 – (0.2)^2 = 100 – 0.04 = 99.96$.
What is the difference between $(x + y + z)^2$ and $x^2 + y^2 + z^2$?
$(x + y + z)^2$ is the square of a trinomial and includes both squared terms and cross-product terms: $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$. It equals $x^2 + y^2 + z^2$ only if the sum of all cross-products $2(xy + yz + zx)$ is zero.
What is the alternative verification form of Identity VIII?
Identity VIII can be rewritten as $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$. This form shows that if $x, y, z$ are positive and real, $x^3 + y^3 + z^3 \ge 3xyz$, with equality holding if and only if $x = y = z$.
Why is $(a – b)^2$ equal to $(b – a)^2$?
Because $(a – b) = -(b – a)$, squaring both sides gives $[-(b – a)]^2 = (-1)^2(b – a)^2 = (b – a)^2$. The square of any real quantity is identical to the square of its additive inverse.
How are algebraic identities used in geometric volume problems?
When a cuboid’s volume is given as a cubic polynomial, factorising that polynomial into three linear factors ($l \times b \times h$) reveals possible algebraic expressions for its length, breadth, and height.
Does $(x + y)^3$ equal $x^3 + y^3$?
No. $(x + y)^3 = x^3 + y^3 + 3xy(x + y)$. Setting $(x + y)^3 = x^3 + y^3$ ignores the two intermediate cross-terms $3x^2y + 3xy^2$, which is a common algebraic error.
How do you evaluate $105 \times 95$ using identities?
Write $105$ as $(100 + 5)$ and $95$ as $(100 – 5)$. Using Identity III: $(100 + 5)(100 – 5) = 100^2 – 5^2 = 10000 – 25 = 9975$.
What are the key presentation steps for 4-mark identity proofs in CBSE exams?
To secure full marks: (1) state the general identity clearly in standard variables, (2) show the step-by-step substitution of terms (with signs enclosed in brackets), (3) simplify each term methodically without skipping steps, and (4) highlight the final factorised or expanded expression clearly.