$ Q(x_{1}) = x_{2}+x_{3} $ , $ Q(x_{2}) = x_{1}+x_{3} $ , $ Q(x_{3}) = x_{1}+x_{2} $ , $ Q(x_{1}x_{2}x_{3}) = 6 $

Question

Let $ P(X) = X^3+7X^2+3X+1 $, with the roots $ x_{1},x_{2},x_{3} \in \mathbb{C} $

Let $ Q $ be a third grade polynomial with the following properties :

$ Q(x_{1}) = x_{2}+x_{3} $

$ Q(x_{2}) = x_{1}+x_{3} $

$ Q(x_{3}) = x_{1}+x_{2} $

$ Q(x_{1}x_{2}x_{3}) = 6 $

Let $ y_{1},y_{2},y_{3} \in \mathbb{C} $ be the roots of $ Q $

$ 1) $ Calculate $y_{1}+y_{2}+y_{3}$

$ 2) $ Calculate $P(y_{1})+P(y_{2})+P(y_{3})$

Both problems could be solved easily with Viete's formulas if we know the coefficients of $ Q $

One easy thing to deduce is that $Q(x_{1}x_{2}x_{3}) = Q(-1)=6 $

But I don't get enough relations to find $a,b,c,d$ in $aX^3+bX^2+cX+d$

Any other ideas?

Answer 1

We have $\sum_i x_i=-7,\,\prod_i x_i = -1$ so $Q(x_i)=-7-x_i,\,Q(-1)=6$. Define $R(x):=Q(x)+x+7$ so $R(x_i)=0,\,R(-1)=12$. Since $P(-1)=4$, $R=3P$ and $Q(x)=3P(x)-x-7=3x^3+21x^2+8x-4$. Thus $\sum_i y_i = -7$ and $\sum_i P(y_i)=\sum_i \frac{y_i+7}{3}=7+\frac{1}{3}\sum_i y_i=\frac{14}{3}$, since each root of $Q$ satisfies $P(y)=\frac{y+7}{3}$.