If x^{4} + frac{1}{x^{4}} = 119,then the valu | vedclass

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(A) Given $x^{4} + \frac{1}{x^{4}} = 119$.We know that $(x^{2} + \frac{1}{x^{2}})^{2} = x^{4} + \frac{1}{x^{4}} + 2$.So,$(x^{2} + \frac{1}{x^{2}})^{2}

Vedclass · Accepted Answer

$\pm 36$

Vedclass · Answer

(A) Given $x^{4} + \frac{1}{x^{4}} = 119$.We know that $(x^{2} + \frac{1}{x^{2}})^{2} = x^{4} + \frac{1}{x^{4}} + 2$.So,$(x^{2} + \frac{1}{x^{2}})^{2} = 119 + 2 = 121$.Taking the square root,$x^{2} + \frac{1}{x^{2}} = 11$ (considering positive value).Now,$(x - \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} - 2 = 11 - 2 = 9$.Thus,$x - \frac{1}{x} = \pm 3$.Case $1$: If $x - \frac{1}{x} = 3$,then $x^{3} - \frac{1}{x^{3}} = (x - \frac{1}{x})^{3} + 3(x - \frac{1}{x}) = 3^{3} + 3(3) = 27 + 9 = 36$.Case $2$: If $x - \frac{1}{x} = -3$,then $x^{3} - \frac{1}{x^{3}} = (-3)^{3} + 3(-3) = -27 - 9 = -36$.Therefore,the value is $\pm 36$.

If $x^{4} + \frac{1}{x^{4}} = 119$,then the value of $x^{3} - \frac{1}{x^{3}}$ is

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