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

(Too long for a comment.)  Building upon this...

Let $\,\sqrt[3]{x+2}=a\,;\;\sqrt[3]{2x-1}=b$

The canonical way to get the polynomial equation in $x$ is to eliminate $a,b$ between:
$$
\begin{cases}
\begin{align}
a^3 &= x+2 \\
b^3 &= 2x-1 \\
a+b +x &= 4
\end{align}
\end{cases}
$$
This is a routine calculation using polynomial resultants, though generally not pretty to do by hand. In this case resultant[ resultant[ a+b+x-4, b^3-2x+1, b ], a^3-x-2, a ] gives $x$ as the root of:
$$
x^9 - 36 x^8 + 585 x^7 - 5589 x^6 + 34317 x^5 - 139563 x^4 + 374328 x^3 - 635553 x^2 + 615033 x - 253503 = 0
$$
Numerically, $x  \;\;\simeq\;\; 1.3254785\dots$
