If $x^3-9x^2+21x-3=0$ then evaluate $x(3-x)^2$ If $x^3-9x^2+21x-3=0$ then evaluate $x(3-x)^2$.
I managed to write the equation as  $(x-3)^3-6(x-3)+6=0$ but could not proceed further.
 A: (First off, I don't see the trick, either, so the following just tackles it the hard way.)
The systematic approach for this would be to let $x (x - 3)^2 = a$, then note that polynomials
$$P(x) = x^3-9x^2+21x-3$$
$$Q(x) = x^3-6x^2+9x-a$$
must have a common root by the given condition and the definition of $a$. This is equivalent to their resultant being $0$, which (after not so pretty calculations, deferred to Wolfram Alpha) reduces to:
$$-a^3+18 a^2-108 a+108 = 0$$
The latter cubic can be solved by Cardano's method for example, with the roots being:
$$a_1 = 6 - 3 \; \sqrt[3]{4}$$
$$a_{2,3} = 6+ 3\;\frac{1 \pm i \sqrt{3}}{\sqrt[3]{2}}$$
The real root $a_1$ is the answer if the question is restricted to real values, all $3$ roots are possible answers in complex numbers.
A: By brute force, you find that the only real solution to 
$ x^3-9x^2+21x-3=0 $ 
is $x = 3-2^{1/3}-2^{2/3}$ and hence 
$$
x(3-x)^2 = 3 (2-2^{2/3})
$$
However, I wonder what is the deeper meaning behind this question?
A: I'm missing "the trick" that would make this easy but you now have a depressed cubic in $y = x-3$:
$$y^3 - 6y + 6 = 0.$$
You can use Vieta's substitution to solve for the real root, back out $x$, and then calculate $x(3-x)^2$.
