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 \(x=1\)

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThanks so much!

Log in to reply

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 \(x=1\)

Log in to reply

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

Log in to reply

1/3

Log in to reply