Complicated algebraic roots

Question

I'm working on a problem:

Suppose that the roots of $$3x^3 +18x^2 +9x -2 = 0$$ are $a$, $b$ and $c$; and the roots of $$x^3 +rx^2 +sx +t = 0$$ are $a+b$, $a+c$ and $b+c$. Determine the cubic equation having a roots of $r$, $s$ and $t$.

My answer is $$x^3 -59x^2 + 876x - 3744.$$ Is my answer wrong?

Answer 1

I think your answer is wrong because $$\sum_{\mathrm{cyc}}(a+b)=2(a+b+c)=-12.$$ $$\sum_{\mathrm{cyc}}(a+b)(a+c)=(a+b+c)^2+ab+ac+bc=36+3=39$$ and $$\prod_{\mathrm{cyc}}(a+b)=(a+b+c)(ab+ac+bc)-abc=-6\cdot3-\frac{2}{3}=-\frac{56}{3}.$$ Can you end it now?

Answer 2

Hint:

Use transformation of equation

As $a+b+c=-\dfrac{18}3,$

let $y=b+c=-6-a\iff a=-6-y$

As $a$ is a root of the given first equation, replace $x$ with $6-y$ and simplify to form a cubic equation in $y$

We can easily discern that $a+b,c+a$ are the rest two roots

Compare it with the given second equation to find the values of $r,s,t$

Can you take it home from here?