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# Can you help me? It is too hard for me to solve!! It has x=1

$$x^{3} -5x^{2} +14x -4 = 6 \times (x^{2} -x +1 )^{1\3}$$

Note by Cong Thanh
1 year, 9 months ago

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Thanks so much!

- 1 year, 6 months ago

I know it's a little late but I'll post my solution, the equation is equivalent to $x^3-5x^2+8x-4+6(x-\sqrt[3]{x^2-x+1})=0$ $\Leftrightarrow (x-1)(x-2)^2+\frac{6(x^3-x^2+x-1)}{x^2+x\sqrt[3]{x^2-x+1}+\sqrt[3]{(x^2-x+1)^2}}=0$ $\Leftrightarrow (x-1)(x-2)^2+\frac{6(x-1)(x^2+1)}{x^2+x\sqrt[3]{x^2-x+1}+\sqrt[3]{(x^2-x+1)^2}}=0$ $\Leftrightarrow (x-1)\bigg[(x-2)^2+\frac{6(x^2+1)}{x^2+x\sqrt[3]{x^2-x+1}+\sqrt[3]{(x^2-x+1)^2}}\bigg]=0$ We see that $$(x-2)^2+\frac{6(x^2+1)}{x^2+x\sqrt[3]{x^2-x+1}+\sqrt[3]{(x^2-x+1)^2}}>0; \forall x\in\mathbb{R}$$ so the only root is $$x=1$$

- 1 year, 6 months ago

May be I can help. But what is the exponent of $$x^{2}-x+1$$ ?

- 1 year, 9 months ago

1/3

- 1 year, 9 months ago