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Question Number 155259 by Fridunatjan08 last updated on 27/Sep/21

$$If:x+\frac{1}{x}=6then{x}^3+\frac{1}{x^3}=?$$

Commented by puissant last updated on 27/Sep/21

$${(x+\frac{1}{x})}^3=6^3=216$$$${(x+\frac{1}{x})}^3=x^3+3x^2\times \frac{1}{x}+3x\times \frac{1}{x^2}+\frac{1}{x^3}$$$$=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=216$$$$\Rightarrow x^3+\frac{1}{x^3}=216-3\times 6=198..$$

Commented by mathdanisur last updated on 27/Sep/21

$$\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{a}\Rightarrow {\mathrm{x}}^3+\frac{1}{{\mathrm{x}}^3}={\mathrm{a}}^3-3\mathrm{a}\Rightarrow 198$$

Answered by Rasheed.Sindhi last updated on 28/Sep/21

$$\mathbb{A}\mathrm{n}\mathbb{A}\mathrm{lternate}\mathbb{W}\mathrm{ay}$$$$\overline{)\begin{array}{c}{a}^3+b^3=(a+b)(a^2-ab+b^2\end{array}}$$$$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2-1+\frac{1}{x^2})$$$$=(x+\frac{1}{x})({(x+\frac{1}{x})}^2-2-1)$$$$=\left(6\right)(6^2-3)=6\times 33=198$$

Commented by peter frank last updated on 28/Sep/21

$$\mathrm{great}$$

Commented by Rasheed.Sindhi last updated on 28/Sep/21

$$\mathrm{Thank}\mathrm{you}!$$

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