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Q6(iv):
Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$

Solution :

Initial Setup & Algebraic Identity

We are tasked with expanding the algebraic expression:

$(x - \frac{2}{3}y)^3$

To expand the cube of a binomial difference, we utilize the standard algebraic identity derived from the Binomial Theorem [where $n=3$]. The identity for the cube of a difference is given by:

$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

x - 2/3 y x - 2/3 y x - 2/3 y Geometric Representation: Volume = (x - 2/3 y)³

Step 1: Variable Mapping and Substitution

By comparing our given expression $(x - \frac{2}{3}y)^3$ with the standard identity $(a - b)^3$, we can establish the following one-to-one mapping of variables:

  • $a = x$
  • $b = \frac{2}{3}y$

Substituting these mapped values into the expansion formula yields:

$(x - \frac{2}{3}y)^3 = (x)^3 - 3(x)^2(\frac{2}{3}y) + 3(x)(\frac{2}{3}y)^2 - (\frac{2}{3}y)^3$

Step 2: Term-by-Term Expansion and Simplification

We will now isolate and simplify each of the four terms generated by the expansion [adhering strictly to the order of operations and the laws of exponents, specifically $(xy)^n = x^n y^n$].

  • First Term ($a^3$):
    $(x)^3 = x^3$

  • Second Term ($-3a^2b$):
    $-3(x)^2(\frac{2}{3}y) = -3 \cdot x^2 \cdot \frac{2}{3}y$
    The scalar $3$ in the numerator and the $3$ in the denominator cancel out:
    $= -2x^2y$

  • Third Term ($+3ab^2$):
    $3(x)(\frac{2}{3}y)^2$
    First, square the term inside the parentheses [$(\frac{2}{3})^2 \cdot y^2 = \frac{4}{9}y^2$]:
    $= 3 \cdot x \cdot \frac{4}{9}y^2$
    Multiply the scalars ($3 \cdot \frac{4}{9} = \frac{12}{9}$), and reduce the fraction by dividing the numerator and denominator by their greatest common divisor, $3$:
    $= \frac{4}{3}xy^2$

  • Fourth Term ($-b^3$):
    $-(\frac{2}{3}y)^3$
    Cube both the coefficient and the variable [$(\frac{2}{3})^3 = \frac{2^3}{3^3} = \frac{8}{27}$]:
    $= -\frac{8}{27}y^3$

Step 3: Final Assembly of the Polynomial

Combine the simplified terms from Step 2 in descending order of the degree of $x$ to form the final expanded polynomial:

$x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$

Final Solution: The expanded form of $(x - \frac{2}{3}y)^3$ is $x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$.


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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