Question
Prove that:
$9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$
$9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$
✓
Answer
We have to prove that $9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$ So,
$9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=3^{2\times\frac{3}{2}}-3\times5^0-\frac{1}{81^{-\frac{1}{2}}}$
$=3^{2\times\frac{3}{2}}-3\times1-\frac{1}{\frac{1}{\sqrt{81}}}$
$9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=3^3-3-\frac{1}{\frac{1}{\sqrt[2]{9\times9}}}$
$=27-3-\frac{1}{\frac{1}{9}}$
$=27-3-1\times\frac{9}{1}$
$=27-12=15$
Hence, $9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$
$9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=3^{2\times\frac{3}{2}}-3\times5^0-\frac{1}{81^{-\frac{1}{2}}}$
$=3^{2\times\frac{3}{2}}-3\times1-\frac{1}{\frac{1}{\sqrt{81}}}$
$9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=3^3-3-\frac{1}{\frac{1}{\sqrt[2]{9\times9}}}$
$=27-3-\frac{1}{\frac{1}{9}}$
$=27-3-1\times\frac{9}{1}$
$=27-12=15$
Hence, $9^{\frac{3}{2}}-3\times5^0-\Big(\frac{1}{81}\Big)^{\frac{-1}{2}}=15$
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