Question

pls tell a relevant answer.

Answer 1

Step-by-step explanation:

We have,

$x = \sqrt[3]{ \sqrt{50} + 7 } - \sqrt[3]{ \sqrt{50} - 7 }$

$\implies \: x^{3} = \bigg(\sqrt[3]{ \sqrt{50} + 7 } - \sqrt[3]{ \sqrt{50} - 7 } \bigg) ^{3}$

$\implies \: x^{3} = \bigg(\sqrt[3]{ \sqrt{50} + 7 } \bigg)^{3} - \bigg( \sqrt[3]{ \sqrt{50} - 7 } \bigg) ^{3} - 3 .\bigg(\sqrt[3]{ \sqrt{50} + 7 } \bigg).\bigg(\sqrt[3]{ \sqrt{50} - 7 } \bigg). \bigg(\sqrt[3]{ \sqrt{50} + 7 } - \sqrt[3]{ \sqrt{50} - 7 } \bigg)$

$\implies \: x^{3} = \sqrt{50} + 7 - \sqrt{50} + 7 - 3 .\bigg(\sqrt[3]{50 - 49} \bigg). (x)$

$\implies \: x^{3} = 14 - 3 x$

$\implies \: x^{3} + 3x - 14 = 0$

$\implies \: x^{3} - 2 {x}^{2} + 2 {x}^{2} - 4x + 7x- 14 = 0$

$\implies \: x^{2}(x - 2) + 2 x(x- 2)+ 7(x- 2) = 0$

$\implies \: (x - 2)( {x}^{2} + 2 x+ 7) = 0$

$\implies \: (x - 2) = 0 \: \: or \: \: ( {x}^{2} + 2 x+ 7) = 0$

But x²+2x+7 is always positive because its a>0 and D<0

So, x=2

Which implies $x^{4}=(2)^{4}=16$