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Question Number 202102 by MATHEMATICSAM last updated on 20/Dec/23

$$\mathrm{If}x=3+2\sqrt{2}\mathrm{then}\frac{x^6+x^4+x^2+1}{x^3}=?$$

Answered by deleteduser1 last updated on 20/Dec/23

$$x+\frac{1}{x}=3+2\sqrt{2}+3-2\sqrt{2}=6$$$$\Rightarrow x^3+x+\frac{1}{x}+\frac{1}{x^3}={(x+\frac{1}{x})}^3-2(x+\frac{1}{x})=204$$

Commented by necx122 last updated on 20/Dec/23

$$Thisiswonderful.Thankyou.$$

Answered by Rasheed.Sindhi last updated on 21/Dec/23

$$AnotherWay$$$$\mathrm{If}x=3+2\sqrt{2}\mathrm{then}\frac{x^6+x^4+x^2+1}{x^3}=?$$$$\frac{1}{x}=\frac{1}{3+2\sqrt{2}}\cdot \frac{3-2\sqrt{2}}{3-2\sqrt{2}}=3-2\sqrt{2}$$$$x+\frac{1}{x}=6$$$$\frac{x^6+x^4+x^2+1}{x^3}=\frac{x^4(x^2+1)+(x^2+1)}{x^3}$$$$=\frac{(x^2+1)(x^4+1)}{x^3}=\frac{x^2+1}{x}\cdot \frac{x^4+1}{x^2}$$$$=(x+\frac{1}{x})(x^2+\frac{1}{x^2})$$$$=(x+\frac{1}{x})({(x+\frac{1}{x})}^2-2)$$$$=\left(6\right)(6^2-2)=6\left(34\right)=204$$

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