# $x_1^3+3x_2+3x_3$ in terms of the roots $x_1$, $x_2$, $x_3$ of $x^3−3x−15=0$

Let $$x_1$$, $$x_2$$, $$x_3$$ be the roots of $$x^3−3x−15=0$$.

Find $$x_1^3+3x_2+3x_3$$.

I tried solving the problem using formulas from Vieta's theorem, but I was unable to find any plausible ways to calculate the end result. Does anyone know how to do this?

$$3(x_1+x_2+x_3)=0$$ since the polynomial has no $$x^2$$ term. Thus, $$x_1^3+3x_2+3x_3=x_1^3-3x_1=15.$$
Let $$a,b,c$$ be the roots of $$x^3 -3x -15=0$$ Let $$I=a^3 + 3b + 3c$$ We have $$a^3 = 3a + 15$$ Since $$a+b+c=- \frac{a_2}{a_3} = 0$$ (Vieta's formulas), $$I= 3a + 15 + 3b + 3c = 15 + 3(a+b+c) = 15$$