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Question Number 162622 by HongKing last updated on 30/Dec/21

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$${\mathrm{x}}^{12}-{15\mathrm{x}}^3+14=0$$

Answered by Ar Brandon last updated on 30/Dec/21

$$x=1$$$$$$$$t^4-15t+14=0,t=x^3$$$$t^4-t-14t+14=0$$$$(t^4-t)-(14t-14)=0$$$$t(t^3-1)-14(t-1)=0$$$$t(t-1)(t^2+t+1)-14(t-1)=0$$$$(t-1)(t^3+t^2+t-14)=0$$$$(t-1)[(t^3-4t)+(t^2+5t-14)]=0$$$$(t-1)[t(t^2-4)+(t-2)(t+7)]=0$$$$(t-1)[t(t-2)(t+2)+(t-2)(t+7)]=0$$$$(t-1)(t-2)(t^2+3t+7)=0$$$$t=1,t=2,t=\frac{-3+i\sqrt{19}}{2},t=\frac{-3-i\sqrt{19}}{2}$$$$x=1,x=\sqrt[3]{2}\mathrm{for}x\in \mathbb{R}$$

Commented by HongKing last updated on 31/Dec/21

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

Answered by aleks041103 last updated on 30/Dec/21

$$t=x^3$$$$t^4-15t+14=0$$$$t^4-t-14t+14=0$$$$t(t^3-1)-14(t-1)=0$$$$t(t-1)(t^2+t+1)-14(t-1)=0$$$$(t-1)(t^3+t^2+t-14)=0$$$$2^3+2^2+2-14=0$$$$\Rightarrow t^3+t^2+t-14=(t-2)(t^2+3t+7)$$$$\Rightarrow t^4-15t+14=(t-1)(t-2)(t^2+3t+7)=0$$$$\Rightarrow t\in \{1,2,\frac{-3\pm \sqrt{9-4.7}}{2}\}\cap \mathbb{R}$$$$\Rightarrow t\in \{1,2\}$$$$\Rightarrow x=1;\sqrt[3]{2}$$$$$$

Commented by HongKing last updated on 31/Dec/21

$$\mathrm{thank}\mathrm{you}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{cool}$$

Answered by MJS_new last updated on 31/Dec/21

$$x^{12}-15x^3+14=0$$$$(x^6-3x^3+2)(x^6+3x^3+7)=0$$$$(x-1)(x^2+x+1)(x^3-2)(x^6+3x^3+7)=0$$$$x_1=1$$$$x_2=2^{1/3}$$$$\mathrm{no}\mathrm{other}\mathrm{real}\mathrm{solutions}$$

Commented by HongKing last updated on 31/Dec/21

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

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