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Question Number 173152 by AgniMath last updated on 07/Jul/22

$$\mathrm{If}x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}\mathrm{then}\mathrm{prove}\mathrm{that}$$$$x^3-6x^2+6x-2=0.$$

Answered by Frix last updated on 07/Jul/22

$$\mathrm{let}x=y^3+y^2+y$$$$y^9+3y^8+6y^7+y^6-6y^5-15y^4-5y^3+6y-2=0$$$$y=2^{1/3}$$$$8+2^{2/3}12+2^{1/3}24+4-2^{2/3}12-2^{1/3}30-10+2^{1/3}6-2=0$$$${\mathrm{ir}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{this}\mathrm{is}\mathrm{true}.$$

Answered by som(math1967) last updated on 07/Jul/22

$$x-2=2^{\frac{2}{3}}+2^{\frac{1}{3}}$$$${(x-2)}^3={(2^{\frac{2}{3}}+2^{\frac{1}{3}})}^3$$$$\Rightarrow x^3-6x^2+12x-8={\left(2^{\frac{2}{3}}\right)}^3+{\left(2^{\frac{1}{3}}\right)}^3$$$$+3.2^{\frac{2}{3}}.2^{\frac{1}{3}}(2^{\frac{2}{3}}+2^{\frac{1}{3}})$$$$\Rightarrow x^3-6x^2+12x-8=4+2+3.2(x-2)$$$$[\because (x-2)=2^{\frac{2}{3}}+2^{\frac{1}{3}}]$$$$\Rightarrow x^3-6x^2+12x-8=6+6x-12$$$$\Rightarrow x^3-6x^2+12x-6x-8+6=0$$$$\therefore x^3-6x^2+6x-2=0$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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