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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?

Thanks in advance!

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    $\begingroup$ I wonder why "precalculus" should be part of a tag that is appropriate for this question. $\endgroup$ Feb 27 at 20:59
  • $\begingroup$ 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... $\endgroup$ Feb 27 at 22:00
  • $\begingroup$ I'm not really sure what's Lagrange multipliers... :( $\endgroup$ Feb 27 at 22:10
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    $\begingroup$ @paul, do you mean Lagrange resolvents? $\endgroup$ Feb 28 at 3:38

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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?

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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.

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

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