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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, 5 months ago

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Thanks so much!

Cong Thanh - 1 year, 2 months ago

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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 \(x=1\)

Gurīdo Cuong - 1 year, 2 months ago

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

Aditya Sky - 1 year, 5 months ago

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1/3

Cong Thanh - 1 year, 5 months ago

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