# 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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Updated on: 10-Oct-2022

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