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Show that if $a$ is a root of $x^3 + x^2 + x – 4$, then $a = a^2 + a – 4$.

Show that if $a$ is a root of $x^3 + x^2 + x – 4$, then $a = a^2 + a – 4$.

I know how to solve this using the method I saw on this website. By using the full terms of the polynomial and multiplying through by $a^2$ I get:
$$
0 = (a^3 – a^2 + a + 4) (a^2 + a – 4)
$$
And then I arrive at:
$$
0 = (a – 4) (a^2 + a + 4)
$$
But I don’t understand how to arrive at this result by multiplying through by $a^2$, which I know is permitted. And I’m also not sure why the $-4$ somehow makes a difference. How do I justify this step?

A:

You can also prove it by algebraic manipulation by multiplying the polynomial by its conjugate $$x^3+x^2+x-4=x^3+x^2+x-4\overline x=(x-\overline x)(x^2+x+4)=a(a-4)$$

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