Cubics

Good question. We know that :

Sum of roots of a cubic $= \frac{\text{-Co-efficient of} \quad {x}^{2}}{\text{Co-efficient of} \quad{x}^{3}}=-3=a+b+c$

Using that we have $P(a)=b+c=(a+b+c)-a=-3-a$

Similarly we have :

$P(b)=-3-b$

$P(c)=-3-c$

$P(a+b+c)=P(-3)=-16$

We assume a polynomial $P(x)+x+3=G(x)$

Hence we have :

$G(a)=G(b)=G(c)=0$

Which implies that $G(x)$ is a cubic polynomial which has roots namely $a,b,c$

We have $G(x)=a({x}^{3}+3{x}^{2}+5{x}+7)$

Now $G(-3)=-16$ , Putting the value we get :

$-16=a(-8)$

$\Rightarrow a=2$

$G(x)=P(x)+x+3=2{x}^{3}+6{x}^{2}+10{x}+14$

$P(x)=2{x}^{3}+6{x}^{2}+9{x}+11$

$\Rightarrow \boxed{P(0)=11}$

Sorry for the lengthy solution

Note that $a + b + c = -3$ by Vieta's. Let $Q(x)$ be the polynomial such that $Q(x) = P(x) + x + 3$ . Note that $Q(x)$ is cubic and has roots $a, b, c$ . So, $Q(x) = C(x^3 + 3x^2 + 5x + 7)$ for some constant $C$ . By the last condition, we have $C = 2$ , so we have $P(0) = Q(0) - 3 = \boxed{11}$ .

Since $a$ , $b$ and $c$ are roots of $x^3+3x^2+5x+7$ , by Vieta's formula, we have $a+b+c = -3$ . And since $P(x)$ is a cubic polynomial and that :

\(\begin{array} {} P(a) = b+c = -a-3 \\ P(b) = c+a = -b-3 \\ P(c) = a+b = -c-3 \end{array} \)

We can conclude that: $P(x) = A(x^3+3x^2+5x+7)-x-3$ , where $A$ is a constant.

Now that:

$P(a+b+c) = P(-3) = A(-27+27-15+7)+3-3 = 8A = -16$

$\Rightarrow A = 2$

$\Rightarrow P(x) = 2(x^3+3x^2+5x+7)-x-3$

$\Rightarrow P(0) = 2(7)-0-3 = \boxed{11}$

I will use Lagrange's Interpolation formula.

By the formula, $P(x)=\sum_{cyclic}((b+c)\frac{(x-b)(x-c)(x-(a+b+c))}{-(a-b)(a-c)(b+c)}) + (-16)\frac{(x-a)(x-b)(x-c)}{(a+b)(b+c)(c+a)}$

Putting $x=0$ and simplifying a bit,we get- $P(0)=(a+b+c)\frac{\sum_{cyclic}ab(a-b)}{(a-b)(a-c)(c-a)}+\frac{16abc}{(a+b)(b+c)(c+a)}=(a+b+c)(1)+\frac{16abc}{(a+b)(b+c)(c+a)}$

We have, $a+b+c=-3,ab+bc+ca=5,abc=-7$ .

Hence, $(a+b)(b+c)(c+a)=(-3-a)(-3-b)(-3-c)=(-3)^3-(-3)^{2}(a+b+c)+(-3)(ab+bc+ca)-abc=(-3)^3-(-3)^{2}(-3)+(-3)(5)-(-7)=-8$

So, $P(0)=-3+\frac{-16.7}{-8}=11$