Thanks so much!
–
Cong Thanh
·
10 months, 3 weeks 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
·
10 months, 3 weeks ago

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May be I can help. But what is the exponent of \(x^{2}-x+1\) ?
–
Aditya Sky
·
1 year, 1 month ago

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TopNewestThanks so much! – Cong Thanh · 10 months, 3 weeks 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 · 10 months, 3 weeks ago

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May be I can help. But what is the exponent of \(x^{2}-x+1\) ? – Aditya Sky · 1 year, 1 month ago

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– Cong Thanh · 1 year, 1 month ago

1/3Log in to reply