Question

The value of $(\frac{1}{216}{)}^{-\frac{1}{3}}$ is
.

Answer 1

$\underset{–}{\text{Need to Find:}}$ $(\frac{1}{216}{)}^{-\frac{1}{3}}$.

To eliminate the negative sign from the exponent, we would take the fraction's (base) reciprocal.

$(\frac{1}{216}{)}^{-\frac{1}{3}}=(216{)}^{\frac{1}{3}}$
$[\because (\frac{a}{b}{)}^{-n}=(\frac{b}{a}{)}^n]$

Now, as we know that $216=6\times 6\times 6=6^3$, we can write:
$(216{)}^{\frac{1}{3}}$
$=(6^3{)}^{\frac{1}{3}}$
$=6^{3\times \frac{1}{3}}$
$(\because (a^m{)}^n=a^{m\times n})$
$=6^1$
$\Rightarrow \underset{–}{(216{)}^{\frac{1}{3}}=(\frac{1}{216}{)}^{-\frac{1}{3}}=6}$

Hence, the value of $(\frac{1}{216}{)}^{-\frac{1}{3}}$ is $6$.