Question

prove that
plz explain it also​

Answer 1

Answer:

LHS = RHS

Step-by-step explanation:

Given,

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

${\huge\red{\Rightarrow}} \: \frac{ {6}^{2} \times {98}^{ \frac{1}{2} } }{ {5}^{2} \times {15}^{ - \frac{4 }{3} } \times {63}^{ \frac{10}{3} - } ( \frac{1}{15} {)}^{ \frac{1}{3} } } = 28 \sqrt{2}$

${\huge\red{\Rightarrow}} \: \frac{ {2}^{3} \times {3}^{3} \times {98}^{ \frac{1}{2} } }{ {5}^{2} \times {5}^{ - \frac{4}{3} } \times {3}^{ \frac{10}{3} - \frac{4}{3} } ( \frac{1}{25} ) ^{ \frac{1}{3} } } = 28 \sqrt{2}$

${\huge\red{\Rightarrow}} \: \frac{ {2}^{2} \times {98}^{ \frac{1}{2} } }{ {5}^{2} \times {5}^{ - \frac{4}{3} } ( \frac{1}{25} ) ^{ \frac{1}{3} } } = 28 \sqrt{2}$

${\huge\red{\Rightarrow}} \: \frac{ {2}^{2} \times 7 \times {2}^{ \frac{1}{2} } }{ {5}^{2} \times \frac{1}{5 \times 5 ^{ \frac{1}{3} } } \times \frac{1}{25 ^{ \frac{1}{3} } } } = 28 \sqrt{2}$

${\huge\red{\Rightarrow}} \: \frac{{2}^{2} \times 7 \times {2}^{ \frac{1}{2} } }{1} = 28 \sqrt{2}$

${\huge\red{\Rightarrow}} \:28 \sqrt{2} = 28 \sqrt{2} \: \: [hence \: proved]$

LHS = RHS. Ans.