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
 A: to solve this equation use $$(a+b)^3=a^3+b^3+3ab(a+b)$$
$$(\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x)^3\\x+2+2x-1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3$$ put $\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x  \\$  so 
$$3x+1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3\\
3x+1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(4-x)=(4-x)^3\\3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(4-x)=(4-x)^3-3x-1\\
27(x+2)(2x-1)(4-x)^3=((4-x)^3-3x-1)^3$$ now you must go to the power of 3 
and solve $ax^9+....=0$ degree=9
but it will be polynomial .
if you solve the eqaution with numerical or graphical method  ,it has a root  $x=1.325$ (the only root ) 
are you sure that type the equation correct ? 
***why there is one real root ? 
  if you take $f(x)=\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x-4 $ so $$f'(x)=\dfrac{1}{3\sqrt[3]{(x+2)^2}}+\dfrac{2}{3\sqrt[3]{(2x-1)^2}}+1 >0$$ so f(x) is strictly increasing $\to $ 
so $$f(x)=0$$ has only one root , It can be checked by 
$$f(1)<0 ,f(1.5)>0 \to x_0 \in (1,1.5)$$ so you can go on 
split $(1,1.5) \to (1,1.25) ,(1.25,1.5)$ 
$$f(1)f(1.25)>0 \\f(1.25)f(1.5)<0 \to x_0 \in (1.25,1.5)$$ and take over 
A: (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, $x  \;\;\simeq\;\; 1.3254785\dots$
A: 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 $a \approx 1.49263$
