Find Polynomial With roots that are the square of the original roots.

Question

I have manipulated the expression to evaluate in such a way that I need to find the value of $a^2b^2+a^2c^2+b^2c^2$, which is the "x" coefficient of the cubic polynomial in which the roots are the squares of the original roots: a,b, and c.

How find this polynomial?

Answer 1

$$x^3+4x=8-3x^2$$

Squaring both sides $$(x(x^2+4))^2=(8-3x^2)^2$$

Replace $x^2$ with $y$

$$y(y+4)^2=(8-3y)^2\iff y^3+y^2(8-9)+y(16+48)-64=0$$ whose roots are $a^2,b^2,c^2$

By Vieta's formula $$a^2+b^2+c^2=\dfrac11$$

$$a^2b^2+b^2c^2+c^2a^2=\dfrac{16+48}1$$

Use $$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)$$

Answer 2

Since the polynomial is monic we have that $$x^3 + 3x^2+4x-8 = (x - a)(x - b)(x -c)$$

Therefore, $$ \begin{array}{r c l} a + b + c & = & -3 \\ ab + ac + bc & = & \;\;\,4 \\ abc & = & \;\;\, 8 \end{array} $$

Square the second equation to get $$ (ab+ac+bc)^2 = (a^2b^2 + a^2c^2+b^2c^2) + 2abc(a+b+c) $$

Hence, the answer is $$ a^2b^2 + a^2c^2+b^2c^2 = 16 + 2\cdot 8 \cdot 3 = 64 $$

Answer 3

Another way, as $x^2(1+x^2) \equiv 6x^2+20x-24 \mod x^3+3x^2+4x-8$, we have $$\sum a^2(1+a^2) = 6\sum a^2+20\sum a - 24\sum 1 = 6\times 1 + 20 \times (-3)+24\times 3 = 18$$ where $\sum$ stands for sum over the roots.