If (x^{3}+frac{1}{x^{3}})=52,then the value o | vedclass

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(A) We know the algebraic identity: $(x+\frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x+\frac{1}{x})$.Given that $x^{3} + \frac{1}{x^{3}} = 52$.Let $

Vedclass · Accepted Answer

$4$

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(A) We know the algebraic identity: $(x+\frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x+\frac{1}{x})$.Given that $x^{3} + \frac{1}{x^{3}} = 52$.Let $y = x + \frac{1}{x}$. Substituting this into the identity,we get:$y^{3} = 52 + 3y$$y^{3} - 3y - 52 = 0$.To solve for $y$,we test integer factors of $52$. For $y = 4$:$4^{3} - 3(4) - 52 = 64 - 12 - 52 = 0$.Since $y = 4$ satisfies the equation,the value of $x + \frac{1}{x}$ is $4$.

If $(x^{3}+\frac{1}{x^{3}})=52$,then the value of $x+\frac{1}{x}$ is

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