Question

4.
Prove that:
give answer please​

Answer 1

$( \frac{ {2}^{ \frac{1}{3} } }{ {2}^{ \frac{ - 1}{3} } } ) {}^{3} + \frac{ {3}^{ \frac{1}{2} } }{ {3}^{ \frac{ - 1}{2} } } = 7$

According to Law of Exponent

$\frac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b}$

Now

$( {2}^{ \frac{1}{3} - ( - \frac{1}{3} )} ) {}^{3} + ( {3}^{ \frac{1}{2} - ( - \frac{ 1}{2}) }) \\ \\ ({2}^{ \frac{1}{3} + \frac{1}{3} } ) {}^{3} + ( {3}^{ \frac{1}{2} + \frac{1}{2} } ) \\ \\ ( {2}^{ \frac{2}{3} } ) {}^{3} + {3}^{ \frac{2}{2} } \\ \\ {2}^{ \frac{2}{3} \times 3 } + 3 \\ \\ {2}^{2} + 3 \\ \\ 4 + 3 \\ \\ = 7$

LHS=7

RHS=7

Hence,

LHS=RHS. (proved)

Answer 2

Answer:

hiiiiiiiii..............