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Prove that:$ \frac{3^{-3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times(15)^{-4 / 3} \times 3^{1 / 3}}=28 \sqrt{2} $
Given:
\( \frac{3^{-3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times(15)^{-4 / 3} \times 3^{1 / 3}}=28 \sqrt{2} \)
To do:
We have to prove that \( \frac{3^{-3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times(15)^{-4 / 3} \times 3^{1 / 3}}=28 \sqrt{2} \).
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 $=\frac{3^{-3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times(15)^{-4 / 3} \times 3^{1 / 3}}$
$=\frac{3^{-3} \times(2\times 3)^{2} \times(2 \times 49)^{\frac{1}{2}}}{5^{2} \times(25)^{\frac{-1}{3}} \times (3\times 5)^{\frac{-4}{3}} \times 3^{\frac{1}{3}}}$
$=\frac{3^{-3} \times 2^{2} \times 3^{2} \times 2^{\frac{1}{2}} \times(7^{2})^{\frac{1}{2}}}{5^{2} \times(5^{2})^{\frac{-1}{3}} \times 3^{\frac{-4}{3}} \times 5^{\frac{-4}{3}} \times 3^{\frac{1}{3}}}$
$=2^{2} \times 2^{\frac{1}{2}} \times 3^{-3+2+\frac{4}{3}-\frac{1}{3}} \times 5^{\frac{2}{3}-2+\frac{4}{3}} \times 7^{1}$
$=4 \sqrt{2} \times 3^{0} \times 5^{0} \times 7$
$=4 \sqrt{2} \times 1 \times 1 \times 7$
$=28 \sqrt{2}$
$=$ RHS
Hence proved.
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