Question

please solve both 1 and 2

Answer 1

Heya there !!!
Here is the answer you were looking for:

Given,

$x + \frac{1}{x} = 7$

(1)

Cubing both the sides,

${(x + \frac{1}{x} )}^{3} = {(7)}^{3} \\ \\ {(x)}^{3} + {( \frac{1}{x} )}^{3} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 343 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(7) = 343 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 21 = 343 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 343 - 21 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 322$

(2)

Again squaring both the sides

${( {x}^{3} + \frac{1}{ {x}^{3} } )}^{2} = {(322)}^{2} \\ \\ {( {x}^{3}) }^{2} + {( \frac{1}{ {x}^{3} } )}^{2} + 2 \times {x}^{3} \times \frac{1}{ {x}^{3} } = 103684 \\ \\ {x}^{6} + \frac{1}{ {x}^{6} } + 2 = 102684 \\ \\ {x}^{6} + \frac{1}{ {x}^{6} } = 102684 - 2 \\ \\ {x}^{6} + \frac{1}{ {x}^{6} } = 102682$

Hope this helps!!!

If you have any doubt regarding to my answer, feel free to ask me in the comment sections ^_^

@Mahak24

Thanks...
☺☺