Roots of polynomials and their formulae relating to coefficients

Question

Write down the cubic equation given that $\alpha + \beta + \gamma = 4$, $\alpha^2 + \beta^2 + \gamma^2 = 66$, and $\alpha^3 + \beta^3 + \gamma^3 = 280$

Ok so, the sum of roots is given and I'm able to use the sum of the roots and the sum of the roots squared to get the sum of the combination of roots, but I'm unable to get the product of roots, because I can't seem to manipulate the sum of the cubes of roots to resemble the sum of roots and sum of squares of roots.

Answer 1

$$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{(\alpha+\beta+\gamma)^2-\alpha^2-\beta^2-\gamma^2}{2}=\frac{16-66}{2}=-25.$$ $$\alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)+3\alpha\beta\gamma.$$ Thus, $$280=64-3\cdot4\cdot(-25)+3\alpha\beta\gamma,$$ which gives $$\alpha\beta\gamma=-28$$ and we got the following equation. $$x^3-4x^2-25x+28=0.$$

Answer 2

Hint, changing to $\;a,b,c\;$ for the roots:

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

$$(a+b+c)^3=a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^3$$

Play with the above and the given data...

Added Or the other way around:

$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$

$$a^3+b^3+c^3=(a+b+c)^3-3a^2(b+c)-3a(b+c)^2$$

Answer 3

Start with $aX^3+bX^2+cX+d = (X-\alpha)(X-\beta)(X-\gamma)$. You get: $ a=1, b = -( \alpha+\beta+\gamma), c=\alpha\beta+\alpha\gamma+\beta\gamma, d=-\alpha\beta\gamma$. These are called the elementary symmetric polynomials (give or take a factor of -1)

Observe that: $ \alpha+\beta+\gamma$

$ \alpha^{2}+\beta^{2}+\gamma^{2}$

and $\alpha^{3}+\beta^{3}+\gamma^{3}$

are all symmetric, and any symmetric polynomial can be expressed using the elementary symmetric polynomials.

Now observe that: $\alpha^2+\beta^2+\gamma^2= (\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\alpha\gamma)$ and that:

$\alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)^3-3(\alpha\beta+\beta\gamma+\alpha\gamma)(\alpha+\beta+\gamma)+3\alpha\beta\gamma$.

You can get these equations on Wolfram using SymmetricReduction[a^3+b^3+c^3, {a,b,c}].

If you plug in the values of the symmetric polynomials you get that:

$b=-4$

$66=16-2c$, so you can get $c$ from this equation.

and finally:

$280=(-4)^3-3c(-4)+3(-d)$. You have c, so you can easily find d, and this gives you the polynomial you are looking for.