If $x^3-9x^2+21x-3=0$ then evaluate $x(3-x)^2$

Question

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.

Answer 1

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 $x(3-x)^2$.

Answer 2

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?

Answer 3

(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.