# Weird Problem on Polynomial Roots

The polynomial $$f(x)=x^3-3x^2-4x+4$$ has three real roots $$r_1$$, $$r_2$$, and $$r_3$$. Let $$g(x)=x^3+ax^2+bx+c$$ be the polynomial which has roots $$s_1$$, $$s_2$$, and $$s_3$$, where \begin{align*} s_1 &= r_1+r_2z+r_3z^2, \\ s_2 &= r_1z+r_2z^2+r_3, \\ s_3 &= r_1z^2+r_2+r_3z, \end{align*}and $$z=\dfrac{-1+i\sqrt3}2$$. Find the real part of the sum of the coefficients of $$g(x)$$.

I know the sum of the coefficients is $$g(1)$$, $$g(x)=(x-s_1)(x-s_2)(x-s_3)$$, and $$z^3=1$$. This means $$s_1z=s_2$$, and $$s_2z=s_3$$. Since $$s_1^3=s_2^3=s_3^3$$, I have $$g(x)=x^3-s_1^3$$. Since the answer is $$g(1)$$, I need to calculate $$1-s_1^3.$$ I expanded $$s_1^3$$ to get $$s_1^3=r_1^3+r_1^2r_2z+3r_1^2r_3z+3r_1r_2^2z^2+6r_1r_2r_3+3r_1r_3^2z+r_2^3+3r_2^2r_3z+3r_2^2r_3z+3r_2r_3^2z^2+r_3^3.$$ I'm pretty sure using Vieta's can finish this, but I'm not sure where else to apply Vieta's other than $$r_1r_2r_3$$. I also tried substituting $$z^2=-z-1$$, but it didn't do much. I also tried using $$(r_1+r_2+r_3)^2$$, but this also failed. Could someone give me some guidance?

• I wonder why "precalculus" should be part of a tag that is appropriate for this question. Feb 27, 2022 at 20:59
• Aren't you re-deriving the Lagrange multipliers approach to solving cubics? My "Algebra" book/notes (freely available on-line) carries this out, as tedious as it may seem... Feb 27, 2022 at 22:00
• I'm not really sure what's Lagrange multipliers... :( Feb 27, 2022 at 22:10
• @paul, do you mean Lagrange resolvents? Feb 28, 2022 at 3:38

Since the answer is g(1), I need to calculate $$1−s_1^3$$.

Well, the real part of $$1−s_1^3$$ is to be calculated, thus we need to calculate $$\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$$ where $$\overline{z}$$ is the complex conjugate of z.

$$\frac{ 1−s_1^3 + \overline{1−s_1^3}}{2}$$ $$\implies\frac{ 2−(s_1^3 + \overline{s_1^3})}{2}$$ $$\implies$$ $$\frac{ 2−(s_1+ \overline{s_1})(s_1^2+\overline{s_1}^2-s_1\overline{s_1})}{2}$$

Also,

$$s_1= r_1+r_2z+r_3z^2$$

$$\overline{s_1}=r_1+r_3z+r_2z^2$$ (as $$z^3=1$$)

On simplifying $$(s_1+ \overline{s_1})(s_1^2+\overline{s_1}^2-s_1\overline{s_1})$$,we obain an expression $$2\displaystyle\sum_{i=1}^{3} r_i^3-3\displaystyle\sum_{1\leq i , j\leq 3,(i≠j) } r_i r_j^2+12\displaystyle\prod_{i=1}^{3} r_i$$ (where $$r_i$$ are roots of f(x)).

Can you proceed further from here?

You know the symmetric polynomials in the roots: $$\sigma_1 = \displaystyle\sum_{i=1}^{3} r_i$$, $$\sigma_2 = \displaystyle\sum_{1\leq i < j\leq 3} r_i r_j$$, and $$\sigma_3 = \displaystyle\prod_{i=1}^{3} r_i$$. Think of combining these into products of degree 3, namely: $$\sigma_1^3, \sigma_1 \sigma_2, \sigma_3$$. Some linear combination of these will give your expression for the real part of $$s_1^3$$. This needs the fact that $$z$$ and $$z^2$$ both have the same real part.

• Wait but how should I deal with the $z$ and $z^2$ that are in some terms but isn't in others? Feb 27, 2022 at 20:46

Your observations are very good and advance quickly to the main task. If we note that $$\ z^2 \ = \ \overline{z} \ \ , \$$ we then have for your sum

$$s_1^3 \ \ = \ \ r_1^3 \ + \ \mathbf{3}r_1^2r_2·z \ + \ 3r_1^2r_3·\overline{z} \ + \ 3r_1r_2^2·z^2 \ + \ 6r_1r_2r_3 \ + \ 3r_1r_3^2·\overline{z}^2 \ + \ r_2^3$$ $$+ \ 3r_2^2r_3·z^2·\overline{z} \ + \ 3r_2r_3^2·z ·\overline{z}^2 \ + \ r_3^3$$ [correcting a couple of oversights] $$= \ \ r_1^3 \ + \ 3 r_1^2r_2·z \ + \ 3r_1^2r_3·\overline{z} \ + \ 3r_1r_2^2·\overline{z} \ + \ 6r_1r_2r_3 \ + \ 3r_1r_3^2· z \ + \ r_2^3$$ $$+ \ 3r_2^2r_3·z \ + \ 3r_2r_3^2· \overline{z} \ + \ r_3^3$$ $$= \ \ r_1^3 \ + \ r_2^3 \ + \ r_3^3 \ + \ 6r_1r_2r_3 \ + \ ( \ 3 r_1^2r_2 \ + \ 3r_1r_3^2 \ + \ 3r_2^2r_3 \ )·z$$ $$+ \ ( \ 3r_1^2r_3 \ + \ 3r_1r_2^2 \ + \ 3r_2r_3^2 \ ) · \overline{z} \ \ ,$$

for which the real part is $$r_1^3 \ + \ r_2^3 \ + \ r_3^3 \ + \ 6r_1r_2r_3 \ - \ \frac32· ( \ r_1^2r_2 \ + \ r_1r_3^2 \ + \ r_2^2r_3 \ + \ r_1^2r_3 \ + \ r_1r_2^2 \ + \ r_2r_3^2 \ ) \ \ .$$

Observing that $$(r_1 + r_2 + r_3)^3 \ \ = \ \ r_1^3 \ + \ r_2^3 \ + \ r_3^3 \ + \ 6r_1r_2r_3$$ $$+ \ 3· ( \ r_1^2r_2 \ + \ r_1r_3^2 \ + \ r_2^2r_3 \ + \ r_1^2r_3 \ + \ r_1r_2^2 \ + \ r_2r_3^2 \ )$$ and $$r_1^2r_2 \ + \ r_1r_3^2 \ + \ r_2^2r_3 \ + \ r_1^2r_3 \ + \ r_1r_2^2 \ + \ r_2r_3^2$$ $$= \ \ (r_1 + r_2 + r_3) · (r_1r_2 \ + \ r_1r_3 \ + \ r_2r_3) \ - \ 3r_1r_2r_3 \ \ ,$$

we obtain $$s_1^3 \ \ = \ \ (r_1 + r_2 + r_3)^3 \ - \ \left(3 + \frac32 \right)·( \ r_1^2r_2 \ + \ r_1r_3^2 \ + \ r_2^2r_3 \ + \ r_1^2r_3 \ + \ r_1r_2^2 \ + \ r_2r_3^2 \ )$$ $$= \ \ (r_1 + r_2 + r_3)^3 \ - \ \frac92 ·[ \ (r_1 + r_2 + r_3) · (r_1r_2 \ + \ r_1r_3 \ + \ r_2r_3) \ - \ 3r_1r_2r_3 \ ] \ \ .$$

Now we can apply the Viete relations to $$\ f(x) \$$ to evaluate this as $$\ 3^3 \ - \ \frac92 ·[ \ 3 · (-4) \ - \ 3·(-4) \ ]$$ $$= \ 27 - 0 \ \ .$$ The real part of the sum of the coefficients of $$\ g(x) \$$ is then $$\ 1 - 27 \ = \ -26 \ \ .$$