Prove that:$ 9^{\frac{3}{2}}-3 \times 5^{0}-\left(\frac{1}{81}\right)^{\frac{-1}{2}}=15 $

Given: 

$9^{\frac{3}{2}}-3 \times 5^{0}-( \frac{1}{81})^{-\frac{1}{2}}=15$.

To do: 

We have to prove that $9^{\frac{3}{2}}-3 \times 5^{0}-( \frac{1}{81})^{-\frac{1}{2}}=15$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

LHS $=9^{\frac{3}{2}}-3 \times 5^{0}-( \frac{1}{81})^{-\frac{1}{2}}$

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

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

$=3^3-3-3^{-4\times-\frac{1}{2}}$

$=3^3-3-3^2$

$=27-3-9$

$=15$

$=$ RHS

Hence proved.