Question

plz !! answer the question​

Answer 1

${\mathtt{\red{Refer~to~the~attachment}}}$

Answer 2

Answer:

a. ${x}^{2} + \frac{1}{ {x}^{2} }$

${x} + \dfrac{1}{ {x}} = 8$

$⟹(x + \dfrac{1}{x} )^{2} = (8)^{2}$

$⟹(x)^{2} + 2 \times x \times \dfrac{1}{x} + ( \dfrac{1}{x} )^{2} = 64$

$⟹ {x}^{2} + 2 + \dfrac{1}{{x}^{2} } = 64$

$⟹ {x}^{2} + \dfrac{1}{ {x}^{2} } = 64 - 2$

Hence , $⟹ {x}^{2} + \dfrac{1}{ {x}^{2} } = 62$

b . ${x}^{4} + \frac{1}{ {x}^{4} }$

as,

${x}^{2} + \dfrac{1}{ {x}^{2} } = 62$

$⟹[ {x}^{2} + \dfrac{1}{ {x}^{2} } ] ^{2} = (62)^{2}$

$⟹({x}^{2} )^{2} + 2 \times {x}^{2} \times \dfrac{1}{ {x}^{2} } + [ \dfrac{1}{ {x}^{2} } ] ^{2} = 3844$

$⟹{x}^{4} + 2 + \dfrac{1}{ {x}^{4} } = 3844$

$⟹ {x}^{4} + \dfrac{1}{ {x}^{4} } = 3844 - 2$

hence, $⟹ {x}^{4} + \dfrac{1}{{x}^{4}} = 3842$