Vieta's Formula and Polynomials

Question

The following below is a question i was asked. The question left me stunned as i was unable to solve it. The question is as follows:

If $A$,$B$ and $C$ are roots of the cubic equation $ax^3+bx^2+cx+d=0$ such that $3c=5a+b+d$ Then the value of $$\frac{(A+B+C)^2(1+A+B+C)-(A^3+B^3+C^3)+ABC(A+B+C+3)}{(A+B+C)}$$ will be:

(a) 3 (b) 5 (c) A^2+3AB - 2 (d) 3ABC-5

Based on my current understanding, i am aware of Vieta's formula for cubics and its results. I would appreciate any help in solving this problem with an explanation.

Answer 1

The polynomial $(x-1)(x-2)(x+(1/3))=x^3-(8/3)x^2+x+(2/3)$ satisfies the hypotheses, and the expression in the question evaluates to $5$, so the answer is (b).

Simpler is the polynomial $x^3-5x^2$ with roots $0,0,5$ and the expression evaluating to $5$.

Answer 2

Expanding and simplifying:

$$ \begin{aligned} \delta & := \frac{(\alpha+\beta+\gamma)^2(\alpha+\beta+\gamma+1)-\left(\alpha^3+\beta^3+\gamma^3\right)+\alpha\,\beta\,\gamma\,(\alpha+\beta+\gamma+3)}{\alpha+\beta+\gamma} \\ & = (\alpha+\beta+\gamma) + 3(\alpha\,\beta+\alpha\,\gamma+\beta\,\gamma) + (\alpha\,\beta\,\gamma) \\ & = -b/a+3\,c/a-d/a \\ & = -b/a+(5\,a+b+d)/a-d/a \\ & = 5 \end{aligned} $$

hence the correct option is b.