Irrational Computation of a Rational Cubic Root. \begin{align}
& x^3+5x-18=0\\
\text{Let } & x=a+b\\
& (a+b)^3+5(a+b)-18 =0\\
& a^3+b^3+(3ab+5)(a+b)-18=0
\end{align}
Taking $a$ and $b$ such that $3ab+5=0$.
We have, $a^3+b^3-18=0$
\begin{align}
& a^3-\frac{5^3}{(3a)^3}-18=0\\
& a^6-18a^3-\frac{125}{27}=0\\
& a^3=9+\sqrt{81+\frac{125}{27}}\\
& a=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}
\end{align}
Therefore, $x=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}$.
I can't solve the last equation without computing the value of the square-root within the cube root. But, by doing that I can only get an approximate value of $x$, not the real value, which is $2$. Is there a way to solve the above equation so that one arrives at the value $2$ and not an approximation?

$$x^3+5x-18=x^3-2x^2+2x^2-4x+9x-18=(x-2)(x^2+2x+9),$$
which gives $x=2$ because $$x^2+2x+9=(x+1)^2+8>0.$$
We need to prove that $$\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=2$$ or
$$9+\sqrt{81+\frac{125}{27}}+9-\sqrt{81+\frac{125}{27}}-2^3+3\cdot2\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}\cdot\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=0$$ or
$$10+6\sqrt[3]{9^2-9^2-\frac{125}{27}}=0,$$ 
which is obvious.
I used $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and the following fact:
$$a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}\sum_{cyc}(a-b)^2=0$$ for $a=b=c$ only.
In our case $a=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}$, $b=\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}$ and $c=-2$.
