Solve for $x \in R$, $x^9 +x^6 = 36$

Solution

Let, $x^3 =t$

We have, $t^3 +t^2 = 36$

$\therefore t^3 +t^2 -36 = 0$

t = 3 satisfies the above equation.

Thus, $t^2 (t-3) + 4t(t-3) + 12(t-3) = 0$

$\therefore (t-3)(t^2+4t+12)=0$

The quadratic $t^2+4t+12=0$ does not yield any real solution since its discriminant is negative.

$\therefore t=3=x^3$

$\therefore x=3^{\frac {1}{3}}$