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 

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

2 equations 3 variables

Given that $x, y, z$ are real, solve (with proof) the system of equations: $$ (x-1)(y-1)(z-1) = xyz - 1$$ $$ (x-2)(y-2)(z-2) = xyz - 2$$ part b) $x,y,z$ are allowed to be complex.

Scroll down for solution.

Let $$P(t) = (x-t)(y-t)(z-t)$$. $x,y,z$ are roots of $P(t) = 0$.

Expand and assume

$$P(t) = c - bt + at^2 - t^3$$

The two equations basically say

$$P(1) = c -1$$

and

$$P(2) = c -2$$

The first gives $a = b$

and second gives $2a - b = 3$ and thus $a=b = 3$

Thus $$P(t) = c - 3t + 3t^2 - t^3 = c -1 - (t^3 -3t^2 + 3t - 1) = (c-1) - (t-1)^3 $$

Thus $x,y,z$ are roots of 

$$(t-1)^3 = c-1$$

For $x,y,z$ to be real, we need $c=1$ and $x=y=z=1$.

If $x,y,z$ are allowed to be complex, pick a random $c$ and $x,y,z$ are $(c-1) w + 1$ where $w$ are the three cuberoots of unity.