If x^{4}+frac{1}{x^{4}}=119 and x1,then find | vedclass

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(C) Given $x^{4}+\frac{1}{x^{4}}=119$.Adding $2$ to both sides,we get $x^{4}+\frac{1}{x^{4}}+2=119+2=121$.This can be written as $(x^{2}+\frac{1}{x^{2

Vedclass · Accepted Answer

$36$

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(C) Given $x^{4}+\frac{1}{x^{4}}=119$.Adding $2$ to both sides,we get $x^{4}+\frac{1}{x^{4}}+2=119+2=121$.This can be written as $(x^{2}+\frac{1}{x^{2}})^{2}=11^{2}$.Taking the square root,$x^{2}+\frac{1}{x^{2}}=11$ (since $x>1$,$x^{2}+\frac{1}{x^{2}}$ must be positive).Now,subtract $2$ from both sides: $x^{2}+\frac{1}{x^{2}}-2=11-2=9$.This simplifies to $(x-\frac{1}{x})^{2}=3^{2}$,so $x-\frac{1}{x}=3$.To find $x^{3}-\frac{1}{x^{3}}$,we use the identity $(x-\frac{1}{x})^{3} = x^{3}-\frac{1}{x^{3}}-3(x-\frac{1}{x})$.Substituting the value $x-\frac{1}{x}=3$,we get $3^{3} = x^{3}-\frac{1}{x^{3}}-3(3)$.$27 = x^{3}-\frac{1}{x^{3}}-9$.Therefore,$x^{3}-\frac{1}{x^{3}} = 27+9 = 36$.

If $x^{4}+\frac{1}{x^{4}}=119$ and $x>1$,then find the positive value of $x^{3}-\frac{1}{x^{3}}$.

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