Question

Evaluate : ${27}^{\frac{2}{3}}+(\frac{81}{16}{)}^{\frac{-1}{4}}$

• A. $9\frac{2}{3}$
• B. $10\frac{2}{3}$
• C. $10$
• D. $\frac{29}{3}$

Answer 1

Answer: (A), (D)

The correct options are
A

$9\frac{2}{3}$

D

$\frac{29}{3}$

${27}^{\frac{2}{3}}+(\frac{81}{16}{)}^{\frac{-1}{4}}$

$=(3^3{)}^{\frac{2}{3}}+(\frac{3^4}{2^4}{)}^{\frac{-1}{4}}$

$=(3^3{)}^{\frac{2}{3}}+((\frac{3}{2}{)}^4{)}^{\frac{-1}{4}}$
$[\because (\frac{a}{b}{)}^m=\frac{a^m}{b^m}]$

$=3^{3\times \frac{2}{3}}+(\frac{3}{2}{)}^{4\times \frac{-1}{4}}$
$[\because (a^m{)}^n=a^{mn}]$

$=3^2+(\frac{3}{2}{)}^{-1}$

$=9+\frac{2}{3}$
​​​​​​​ $[\because (a{)}^{-m}=\frac{1}{a^m}]$

$=\frac{29}{3}$

$=9\frac{2}{3}$