Cute problem from IIT JEE

If $r,s,t$ are roots (you can assume they are all real) of $$x^3 + 3x^2 - 24x + 1 = 0$$ Find $$\sqrt[3]{r} + \sqrt[3]{s} + \sqrt[3]{t}$$ If you know which year this is from, please let me know.

Scroll down for solution (and something extra).

Let $P(x) = x^3 + 3x^2 - 24x + 1$.

Note that $P(0) > 0, P(1) < 0$ and $P(24) > 0$ and so has at least two real roots. Which implies all roots are real, so we can freely take cube roots.

Now 

$$x^3 +3x^2 - 24x + 1 = x^3 + 3x^2 + 3x + 1 - 27x = (x+1)^3 - 27x$$ .

Thus if $a$ is a root of $P$, then

$$(a+1)^3 = 27a \implies 3a^{1/3} = (a + 1)$$

Thus the sum of cuberoots of P is $(r+s+t + 3)/3 = 0$.

For the extra:

We will in fact show that $\sqrt[3]{r},\sqrt[3]{s},\sqrt[3]{t}$ are roots of $x^3 - 3x + 1 = 0$.

Let $d=\sqrt[3]{r},e=\sqrt[3]{s},f=\sqrt[3]{t}$ be roots of $Q(x) = x^3 + ax^2 + bx + 1$

Thus $Q(x) = (x-d)(x-e)(x-f)$

Now if $w$ is a cuberoot of unity, then

$$Q(x)Q(wx)Q(w^2x) = (x^3 - d^3)(x^3-e^3)(x^3 - f^3) = P(x^3) = x^9 + 3x^6 -24x^3 + 1$$

With slightly tedious algebra, we see that

$$Q(x)Q(wx)Q(w^2x) = x^9 + (a^3 - 3ab + 3)x^6 + (b^3 - 3ab + 3)x^3 + 1$$

Thus we get

$$a^3 - 3ab = 0$$

$$b^3 - 3ab = -27$$  

From the first equation either $a=0$ (in which case $b = -3$) or $a^2 = 3b$.

If $a \neq 0$, putting $b = a^2/3$ in the second results in a quadratic (in $a^3$) with complex roots. Since $a$ needs to be real (sum of real numbers) we see that $a = 0, b = -3$ and hence the cuberoots are roots of 

$$x^3 - 3x + 1 = 0$$