Question

x-1/x =3 then x⁴+1/x⁴​

Answer 1

$x-\dfrac{1}{x}=3\\\\x^2-2+\dfrac{1}{x^2}=9\\\\x^2+\dfrac{1}{x^2}=11\\\\x^4+2+\dfrac{1}{x^4}=121\\\\x^4+\dfrac{1}{x^4}=119$

Answer 2

Question :-

$\sf{if \: x - \frac{1}{x} = 3 \: then \: find \: {x}^{4} + \frac{1}{ {x}^{4} } }$

Answer :-

$\implies \: \boxed{ \sf{ {x}^{4} + \frac{1}{ {x}^{4} } = 119}} \:$

To Find :-

$\implies \sf{ {x}^{4} + \frac{1}{ {x}^{4} } }$

Formula used :-

$\rightarrow \star \: \: \sf{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy} \\ \rightarrow \star \: \sf{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

Solution :-

According to the question,

$\rightarrow \: \sf{x - \frac{1}{x} = 3} \\ \\ \sf{squaring \: on \: both \: sides \: } \\ \\ \rightarrow \sf{ {x}^{2} + \frac{1}{ {x}^{2} } - 2 \: x \times \frac{1}{x} = 9} \\ \\ \rightarrow \sf{ {x}^{2} + \frac{1}{ {x}^{2} } = 11} \\ \\ \text{again \: squaring \: on \: both \: sides \: } \\ \\ \rightarrow \sf{ {x}^{4} + \frac{1}{ {x}^{4} } + 2 {x}^{2} \times \frac{1}{ {x}^{2} } = 121} \\ \\ \rightarrow \boxed{ \sf{ {x}^{4} + \frac{1}{ {x}^{4} } = 119}}$

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