Question

Evaluate: $(\frac{x}{2}+2y)(\frac{x^2}{4}-xy+4y^2)$

• A. $\frac{x^3}{11}+8y^3$
• B. $\frac{x^3}{8}+8y^3$
• C. $\frac{x^3}{14}+8y^3$
• D. $\frac{x^3}{7}+8y^3$

Answer 1

Answer: (B)

$\frac{x^3}{8}+8y^3$

We know,
$(a^3+b^3)=a^3+a^{2}b-a^{2}b-a{b}^2+a{b}^2+b^3$
$=(a^3+a^{2}b)-(a^{2}b+a{b}^2)+(a{b}^2+b^3)$
$=a^2(a+b)-ab(a+b)+b^2(a+b)$
$=(a+b)(a^2-ab+b^2)$
Now, Given $(\frac{x}{2}+2y)(\frac{x^2}{4}-xy+4y^2)$
$=(\frac{x}{2}+2y)\left[\right(\frac{x}{2}{)}^2-\left(\frac{x}{2}\right)\left(y\right)+\left(2y{)}^2\right]$
$=(\frac{x}{2}{)}^3+(2y{)}^3$
$=\frac{x^3}{8}+8y^3$