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.
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