help

pls try to have one question per post :)

1.

Let the polynomial be $f(x).$

It is useful to find a polynomial with roots of $a^2,b^2,c^2$ so we can quickly solve it using Vieta's.

Notice that $f(\sqrt{x})$ has roots of $a^2, b^2, c^2$, but only if that root is positive (specifically if the real part is positive). That is because $\sqrt{x}=k$ for some constant where the real part is negative has no solutions.

Also, notice that $f(-\sqrt{x})$ has roots of $a^2, b^2, c^2$, but only of the real part of that root is negative. That means that $f(\sqrt{x})\cdot f(-\sqrt{x})$ will ensure that it has roots of $a^2, b^2, c^2$, regardless of if the roots are positive or negative. 

Also note that since the polynomial $f(\sqrt{x})\cdot f(-\sqrt{x})$ is even, it can be expressed as a polynomial in terms of $\sqrt{x}^2=x$.

Let g(x) be the polynomial with roots of $a^2,b^2,c^2$. We can see that the polynomial, after some simplifying, is equal to:

$-x^3 + 2 x^2 - x + 16$

Note that $(a^2 - 1)(b^2 - 1)(c^2 - 1)=a^2 b^2 c^2 - a^2b^2 - a^2 c^2 - b^2 c^2 + a^2 + b^2 + c^2 - 1$. By Vieta's, $a^2b^2c^2=-16, a^2b^2+a^2c^2+b^2c^2=-1, a^2+b^2+c^2=-2$, and the rest can be solved easily.

2.

This problem can be solved quickly using polynomial interpolation using remainder theorem as follows:

Notice that $P(x)-x=0$ for $x=1, 2, 3, 4$, which means that $P(x)-x=A(x-1)(x-2)(x-3)(x-4)$ for some constant A. Rearranging, we get that $P(x)=x+A(x-1)(x-2)(x-3)(x-4)$. We can solve for A by substituting 5 for x:

$5+A(5-1)(5-2)(5-3)(5-4)=125\\ A(4)(3)(2)=120\\ A=5$

So the polynomial is equal to $5+5(x-1)(x-2)(x-3)(x-4)$.

Substitute x=6 to get the answer.