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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?

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$3(x_1+x_2+x_3)=0$ since the polynomial has no $x^2$ term. Thus, $$ x_1^3+3x_2+3x_3=x_1^3-3x_1=15. $$

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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), $$ I= 3a + 15 + 3b + 3c = 15 + 3(a+b+c) = 15 $$

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