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Q9(i):
Verify : (i) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
Solution :
Initial Setup & Algebraic Objective
We are required to rigorously verify the standard algebraic identity for the sum of two cubes. The equation provided is:
$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
To prove this equivalence, we will designate the expressions as follows:
- Left Hand Side (LHS): $x^3 + y^3$
- Right Hand Side (RHS): $(x + y)(x^2 - xy + y^2)$
Our objective is to expand the RHS and demonstrate that it simplifies exactly to the LHS.
Step 1: Applying the Distributive Property
We begin by expanding the RHS using the distributive property of multiplication over addition [Algebraic Axiom: $a(b + c + d) = ab + ac + ad$]. We must distribute both terms of the binomial $(x + y)$ across the trinomial $(x^2 - xy + y^2)$.
$RHS = x(x^2 - xy + y^2) + y(x^2 - xy + y^2)$
Step 2: Term-by-Term Algebraic Expansion
Next, we perform the scalar multiplication for each distributed component, applying the laws of exponents [Specifically, the Product Rule: $a^m \cdot a^n = a^{m+n}$].
Expanding the first cluster:
- $x \cdot (x^2) = x^3$
- $x \cdot (-xy) = -x^2y$
- $x \cdot (y^2) = xy^2$
Resulting polynomial segment: $x^3 - x^2y + xy^2$
Expanding the second cluster:
- $y \cdot (x^2) = x^2y$
- $y \cdot (-xy) = -xy^2$
- $y \cdot (y^2) = y^3$
Resulting polynomial segment: $x^2y - xy^2 + y^3$
Step 3: Aggregating and Simplifying Like Terms
We now combine both expanded segments into a single, unified polynomial expression:
$RHS = x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3$
To simplify, we group the "like terms" (terms containing the exact same variables raised to the exact same powers) using the Commutative Property of Addition:
$RHS = x^3 + (-x^2y + x^2y) + (xy^2 - xy^2) + y^3$
| Like Term Group | Operation | Result (Additive Inverse) |
|---|---|---|
| $x^2y$ terms | $-x^2y + x^2y$ | $0$ |
| $xy^2$ terms | $xy^2 - xy^2$ | $0$ |
Because these pairs are additive inverses of one another, they sum to zero and cancel out completely.
Step 4: Final Equivalence
Substituting the zeroed terms back into the equation yields:
$RHS = x^3 + 0 + 0 + y^3$
$RHS = x^3 + y^3$
Comparing this result to our initial setup, we observe that the simplified RHS is identical to the LHS.
$LHS = RHS$
Final Solution: The identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ is rigorously verified.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4
- Q1(i): Use suitable identities to find the following products: (i) $(x + 4) (x + 10)$
- Q1(ii): Use suitable identities to find the following products: (ii) $(x + 8) (x – 10)$
- Q1(iii): Use suitable identities to find the following products: (iii) $(3x + 4) (3x – 5)$
- Q1(iv): Use suitable identities to find the following products: (iv) $(y^2 + \frac{3}{2}) (y^2 – \frac{3}{2})$
- Q1(v): Use suitable identities to find the following products: (v) $(3 – 2x) (3 + 2x)$
- Q10(i): Factorise each of the following: (i) $27y^3 + 125z^3$ [Hint : See Question 9.]
- Q10(ii): Factorise each of the following: (ii) $64m^3 – 343n^3$ [Hint : See Question 9.]
- Q11: Factorise : $27x^3 + y^3 + z^3 – 9xyz$
- Q12: Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
- Q13: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
- Q14(i): Without actually calculating the cubes, find the value of each of the following: (i) $(–12)^3 + (7)^3 + (5)^3$
- Q14(ii): Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$
- Q15(i): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : $25a^2 – 35a + 12$
- Q15(ii): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$
- Q16(i): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : $3x^2 – 12x$
- Q16(ii): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : $12ky^2 + 8ky – 20k$
- Q2(i): Evaluate the following products without multiplying directly: (i) $103 \times 107$
- Q2(ii): Evaluate the following products without multiplying directly: (ii) $95 \times 96$
- Q2(iii): Evaluate the following products without multiplying directly: (iii) $104 \times 96$
- Q3(i): Factorise the following using appropriate identities: (i) $9x^2 + 6xy + y^2$
- Q3(ii): Factorise the following using appropriate identities: (ii) $4y^2 – 4y + 1$
- Q3(iii): Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$
- Q4(i): Expand each of the following, using suitable identities: (i) $(x + 2y + 4z)^2$
- Q4(ii): Expand each of the following, using suitable identities: (ii) $(2x – y + z)^2$
- Q4(iii): Expand each of the following, using suitable identities: (iii) $(–2x + 3y + 2z)^2$
- Q4(iv): Expand each of the following, using suitable identities: (iv) $(3a – 7b – c)^2$
- Q4(v): Expand each of the following, using suitable identities: (v) $(–2x + 5y – 3z)^2$
- Q4(vi): Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$
- Q5(i): Factorise: (i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
- Q5(ii): Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
- Q6(i): Write the following cubes in expanded form: (i) $(2x + 1)^3$
- Q6(ii): Write the following cubes in expanded form: (ii) $(2a – 3b)^3$
- Q6(iii): Write the following cubes in expanded form: (iii) $(\frac{3}{2}x + 1)^3$
- Q6(iv): Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$
- Q7(i): Evaluate the following using suitable identities: (i) $(99)^3$
- Q7(ii): Evaluate the following using suitable identities: (ii) $(102)^3$
- Q7(iii): Evaluate the following using suitable identities: (iii) $(998)^3$
- Q8(i): Factorise each of the following: (i) $8a^3 + b^3 + 12a^2b + 6ab^2$
- Q8(ii): Factorise each of the following: (ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
- Q8(iii): Factorise each of the following: (iii) $27 – 125a^3 – 135a + 225a^2$
- Q8(iv): Factorise each of the following: (iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
- Q8(v): Factorise each of the following: (v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
- Q9(ii): Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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