If x1 and x^{2}+frac{1}{x^{2}}=83, then x^{3 | vedclass

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(C) Given $x^{2}+\frac{1}{x^{2}}=83$.We know that $(x-\frac{1}{x})^{2} = x^{2}+\frac{1}{x^{2}}-2$.Substituting the value,we get $(x-\frac{1}{x})^{2} =

Vedclass · Accepted Answer

$756$

Vedclass · Answer

(C) Given $x^{2}+\frac{1}{x^{2}}=83$.We know that $(x-\frac{1}{x})^{2} = x^{2}+\frac{1}{x^{2}}-2$.Substituting the value,we get $(x-\frac{1}{x})^{2} = 83-2 = 81$.Since $x>1$,$x-\frac{1}{x}$ must be positive,so $x-\frac{1}{x} = \sqrt{81} = 9$.Now,using the identity $(x-\frac{1}{x})^{3} = x^{3}-\frac{1}{x^{3}}-3(x-\frac{1}{x})$.Substituting the values,$9^{3} = x^{3}-\frac{1}{x^{3}}-3(9)$.$729 = x^{3}-\frac{1}{x^{3}}-27$.Therefore,$x^{3}-\frac{1}{x^{3}} = 729+27 = 756$.

If $x>1$ and $x^{2}+\frac{1}{x^{2}}=83,$ then $x^{3}-\frac{1}{x^{3}}$ is

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