Question

$\huge \pink{ \boxed{ \sf{ \green{Question}}}}$

Prove that:

$\huge \frac{ {3}^{ - 3} \times {6}^{2} \times \sqrt{98} }{ {5}^{2} \times \sqrt[3]{ \frac{1}{25} } \times ( {15})^{ \frac{ - 4}{3}} \times {3}^{ \frac{1}{3} } } = 28 \sqrt{2}$
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Answer 1

$\pink{\begin{gathered} \frac{{3}^{ - 3} \times {6}^{2} \times \sqrt{98 }}{{5}^{2} \times \sqrt[3]{ \frac{1}{25} } \times {15}^{ \frac{ - 4}{3} } \times {3}^{ \frac{1}{3}}} \\ \\ = \frac{{3}^{ - 3} \times {6}^{2} \times 7\sqrt{2 }}{{5}^{2} \times {5}^{ \frac{ - 2}{3} } \times {5}^{ \frac{ - 4}{3} }\times {3}^{ \frac{ - 4}{3} } \times {3}^{ \frac{1}{3}}} \\ \\ = \frac{{3}^{ - 3} \times {3}^{2} \times {2}^{2} \times {2}^{ \frac{1}{2} } \times 7}{{5}^{2} \times {5}^{ \frac{ - 2}{3} } \times {5}^{ \frac{ - 4}{3} }\times {3}^{ \frac{ - 4}{3} } \times {3}^{ \frac{1}{3}}} \\ \\ = {3}^{( - 3 + 2 + \frac{4}{3} - \frac{1}{3}) } \times {2}^{(2 + \frac{1}{2}) } \times 7 \times {5}^{ - 2 + \frac{2}{3} + \frac{4}{3} } \\ \\ = {3}^{0} \times {2}^{ \frac{5}{2} } \times 7 \times {5}^{0} \\ \\ = 4 \sqrt{2} \times 7 \\ \\ = 28 \sqrt{2} \end{gathered} </p><p>}$

Answer 2

Refer this attachment

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