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Question Number 66197 by Rasheed.Sindhi last updated on 11/Aug/19

$$\mathrm{If}\mathrm{x}+\frac{1}{\mathrm{x}}=1,\mathrm{prove}\mathrm{that}:$$$${\mathrm{x}}^{\mathrm{n}}+{\mathrm{x}}^{\mathrm{n}-2}+{\mathrm{x}}^{\mathrm{n}-4}=0$$

Commented by mr W last updated on 10/Aug/19

$$\mathcal{VERY}\mathcal{NICE}!$$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

$${\mathrm{x}}^{\mathrm{n}}+{\mathrm{x}}^{\mathrm{n}-2}+{\mathrm{x}}^{\mathrm{n}-4}$$$$={\mathrm{x}}^{\mathrm{n}-2}({\mathrm{x}}^2+1+\frac{1}{{\mathrm{x}}^2})$$$$={\mathrm{x}}^{\mathrm{n}-2}({\mathrm{x}}^2+2+\frac{1}{{\mathrm{x}}^2}-1)$$$$={\mathrm{x}}^{\mathrm{n}-2}({(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-1)$$$$={\mathrm{x}}^{\mathrm{n}-2}({\left\{1\right\}}^2-1)$$$$={\mathrm{x}}^{\mathrm{n}-2}.0$$$$=0$$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

$$\mathcal{T}hank\mathcal{S}sfor\mathcal{E}ncouraging\mathcal{S}ir!$$

Answered by som(math1967) last updated on 10/Aug/19

$$x+\frac{1}{x}=1$$$$\frac{x^2+1}{x}=1$$$$x^2-x+1=0\Rightarrow (x+1)(x^2-x+1)=0$$$$\Rightarrow x^3+1=0\therefore x^3=-1\Rightarrow x^3.x=-x$$$$\therefore x^4=-x$$$$Now{x}^n+x^{n-2}+x^{n-4}$$$$=x^n(1+x^{-2}+x^{-4})$$$$=x^n(1+\frac{1}{x^2}+\frac{1}{x^4})=x^n\left(\frac{x^4+x^2+1}{x^4}\right)$$$$=x^n\left(\frac{x^2-x+1}{x^4}\right)★$$$$=x^n\left(\frac{0}{x^4}\right)=0★[★$$$$★x^4=-x$$$$★★x^2-x+1=0$$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

$$\mathcal{TH}anks\mathcal{S}ir!$$

Answered by mr W last updated on 10/Aug/19

$$x\ne 0$$$$x+\frac{1}{x}=1$$$$\Rightarrow x^2-x+1=0$$$$\Rightarrow x^2=x-1$$$$\Rightarrow x^4={(x-1)}^2=x^2-2x+1$$$$\Rightarrow x^4+x^2+1=2x^2-2x+2$$$$\Rightarrow x^4+x^2+1=2(x^2-x+1)$$$$\Rightarrow x^4+x^2+1=0$$$$\Rightarrow x^{n-4}(x^4+x^2+1)=0$$$$\Rightarrow x^n+x^{n-2}+x^{n-4}=0$$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

$$\mathbb{T}\mathrm{h}\alpha \mathrm{n}\mathbb{k}\mathrm{s}\mathrm{Sir}!$$

Answered by gunawan last updated on 11/Aug/19

$$x^2+x^{-2}+2=1$$$$x^2+x^{-2}+1=0$$$$$$$$x^n+x^{n-2}+x^{n-4}=0$$$$x^{n-2}(1+x^2+x^{-2})$$$$=x^{n-2}\times 0=0$$

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