Please help me solve this question in a easy way:
$$ \sqrt[3]{a + \frac{a+8}{3}\sqrt{\frac{a-1}{3}}} + \sqrt[3]{a - \frac{a+8}{3}\sqrt\frac{a-1}{3}} = m $$
Find $m^3$.
(The answer is $8$.)
I tried solving the problem by substituting $x = (a-1/3)^{1/2}$ then used the identity $(a+b)^3 = a^3 + b^3 +3ab(a+b)$ this formed a cubic polynomial. But then I could not proceed. I was unable to solve ab. So please help in any way possible.
The domain gives $a\geq1$.
For $a=1$ we obtain $m=2$ and $m^3=8$.
Let $a>1$.
Thus, since $$a^3+b^3+c^3-3abc=(a+b+c)\sum_{cyc}(a^2-bc)=\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2,$$ we see that $$a^3+b^3+c^3=3abc,$$ when $$a+b+c=0$$ or $$\sum_{cyc}(a-b)^2=0.$$ The last is possible for $a=b=c$.
But for $a>1$ we see that $$ \sqrt[3]{a + \frac{a+8}{3}\sqrt{\frac{a-1}{3}}}\neq\sqrt[3]{a - \frac{a+8}{3}\sqrt\frac{a-1}{3}},$$ which says $$\sqrt[3]{a + \frac{a+8}{3}\sqrt{\frac{a-1}{3}}}+\sqrt[3]{a - \frac{a+8}{3}\sqrt\frac{a-1}{3}}=m$$ it's $$\left(\sqrt[3]{a + \frac{a+8}{3}\sqrt{\frac{a-1}{3}}}\right)^3+\left(\sqrt[3]{a - \frac{a+8}{3}\sqrt\frac{a-1}{3}}\right)^3-m^3+$$ $$+3m\sqrt[3]{a + \frac{a+8}{3}\sqrt{\frac{a-1}{3}}}\cdot\sqrt[3]{a - \frac{a+8}{3}\sqrt\frac{a-1}{3}}=0$$ or $$2a-m^3+3m\sqrt[3]{a^2-\frac{(a+8)^2(a-1)}{27}}=0$$ or $$2a-m^3+m\sqrt[3]{-a^3+12a^2-48a+64}=0$$ or $$2a-m^3+m(4-a)=0$$ or $$m^3-4m+(m-2)a=0$$ or $$(m-2)(m^2+2m+a)=0$$ or $$m=2,$$ which gives $$m^3=8$$