<s>[INST] Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$ Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$

The root of equation very bad
My try 1: 
Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$
Have: $a+b=4-x$ 
=>Root of a system of equation bad, too
My try 2:
Use quality $\sqrt[3]{a}\pm \sqrt[3]{b}=\frac{a\pm b}{\sqrt[3]{a^2}\mp \sqrt[3]{ab}+\sqrt[3]{b^2}}$ :
$\sqrt[3]{x+2}-ax+\sqrt[3]{2x-1}-bx=4-x$
Need find $ax$, $bx$ but it's very bad, too
 [/INST] If we set $a = \sqrt[3]{x+2}$ and $b = \sqrt[3]{2x-1}$, since $a + b = 4 - x = 1 + (3 - x)$, we obtain that
\begin{cases}
a^{3} - b^{3}  = a + b - 1\\
b^{3} = 2a^{3} - 5
\end{cases}
Hence we conclude that $b = 6 - a - a^{3}$, from whence we have the equation
\begin{align*}
(6 - a - a^{3})^{3} = 2a^{3} - 5
\end{align*}
Which can be solved numerically. Wolfram, for example, gives $a \approx 1.49263$
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