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

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 
