Question

Find the cube of $\frac{x}{4}+\frac{4y}{3}$

• A. $\frac{x^3}{64}+\frac{64y^3}{27}+3xy(\frac{x}{4}+\frac{4y}{3})$
• B. $\frac{x^3}{64}+\frac{64y^3}{27}+xy(\frac{x}{4}+\frac{4y}{3})$
• C. $\frac{x^3}{64}+\frac{64y^3}{27}-xy(\frac{x}{4}+\frac{4y}{3})$
• D. $\frac{x^3}{64}+\frac{64y^3}{27}-3xy(\frac{x}{4}+\frac{4y}{3})$

Answer 1

The correct option is B

$\frac{x^3}{64}+\frac{64y^3}{27}+xy(\frac{x}{4}+\frac{4y}{3})$

${(\frac{x}{4}+\frac{4y}{3})}^3={\left(\frac{x}{4}\right)}^3+{\left(\frac{4y}{3}\right)}^3+3\times \frac{x}{4}\times \frac{4y}{3}\times (\frac{x}{4}+\frac{4y}{3})$

$=\frac{x^3}{64}+\frac{64y^3}{27}+xy(\frac{x}{4}+\frac{4y}{3})$