<s>[INST] Show that $c$ is a root of the equation $6x^3-6sx^2+3(s^2-t)x+3st-s^3-2u=0$ 
(a) Let $p,q$ and $r$ be real numbers. Given that there are numbers $a$ and $b$ such that $$a+b=p, ~~~ a^2+b^2=q, ~~~ a^3+b^3=r~~~~~~~~~~~~~~~(*)$$
Show that $3pq - p^3=2r$

Workings: $3pq-p^3 = 3(a+b)(a^2+b^2)-(a+b)^3 = 2a^3+2b^3=2r$

(b) Conversely, you are given that the real numbers $p,q$ and $r$ satisfy $3pq-p^3=2r$. By considering the equation $2x^2-2px+(p^2-q)=0$, show that there exist numbers $a$ and $b$ such that the three equations $(*)$ hold

Workings: Suppose that the roots of the quadratic are $$2x^2-2px+(p^2-q)=0 \iff (x-a)(x-b)=0\\ \therefore a+b=p, ~~~ ab=\frac{p^2-q}{2} \\ a^2+b^2=(a+b)^2-2ab \implies a^2+b^2=q \\ a^3+b^3 = (a+b)^3-3ab(a+b) \implies a^3+b^3= p^3-3(\frac{p^2-q}{2})(p)=\frac{1}{2}(3pq-p^3)=r$$
Hence, if $a,b$ satisfy the quadratic equation, then we have shown that it also satisfies the three equations in $(*)$

(c) Let $s,t,u$ and $v$ be real numbers. Given that there distinct numbers such that $a,b$ and $c$ such that $$a+b+c=s,~~~ a^2+b^2+c^2=t,~~~a^3+b^3+c^3=u,~~~ abc=v$$
Show, using part (a), that $c$ is a root of the equation $$6x^3-6sx^2+3(s^2-t)x+3st-s^3-2u=0$$

I am stuck on this part of the problem. I used the same approach as (a) to solve this but scored 0 marks. As I didn't use part (a). Could I have an explanation on how to show this using part (a)?
 [/INST] In order to make use of (a), note that
$$\begin{aligned}
a+b&=s-c=:p\\
a^2+b^2&=t-c^2=:q\\
a^3+b^3&=u-c^3=:r
\end{aligned}$$
so that
$$ 3pq - p^3=2r$$
translates to
$$3(s-c)(t-c^2)-(s-c)^3=2(u-c^3). $$
Expand and rearrange this.
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