Question

Expand ${(\frac{1}{2}p+\frac{1}{3}q)}^3$

• A. $\frac{1}{8}{p}^3+\frac{1}{27}{q}^3+\frac{1}{4}{p}^{2}q-\frac{1}{6}p{q}^2$
• B. $\frac{1}{8}{p}^3+\frac{1}{27}{q}^3+\frac{1}{4}{p}^{2}q+\frac{1}{6}p{q}^2$
• C. $\frac{1}{8}{p}^2+\frac{1}{27}{q}^3+\frac{1}{4}{p}^{2}q+\frac{1}{6}p{q}^2$
• D. $\frac{1}{8}{p}^2+\frac{1}{27}{q}^3-+\frac{1}{4}{p}^{2}q+\frac{1}{6}p{q}^2$

Answer 1

Answer: (B)

$\frac{1}{8}{p}^3+\frac{1}{27}{q}^3+\frac{1}{4}{p}^{2}q+\frac{1}{6}p{q}^2$

${(\frac{1}{2}p+\frac{1}{3}q)}^3$
Using the formula,$(a+b{)}^3=a^3+b^3+3ab(a+b)$
$=(\frac{1}{2}p{)}^3+(\frac{1}{3}q{)}^3+3\times \left(\frac{1}{2}p\right)\times \left(\frac{1}{3}q\right)(\frac{1}{2}p+\frac{1}{3}q)$
$=\frac{1}{8}{p}^3+\frac{1}{27}{q}^3+\frac{1}{4}{p}^{2}q+\frac{1}{6}p{q}^2$