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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 $(x^2 - xy + y^2)$.

$RHS = x(x^2 - xy + y^2) + y(x^2 - xy + y^2)$

( x + y ) ( x² - xy + y² ) Distributing x Distributing y

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


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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