6.364 1. Let $x_{1}, x_{2}, x_{3}$ be the roots of the polynomial $a x^{3}+b x^{2}+c x+d$.

## Use the identity

$a x^{3}+b x^{2}+c x+d=a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)$ to obtain formulas relating the roots and coefficients of the given polynomial.

2. Find the roots $x_{1}, x_{2}, x_{3}$ of the equation

$8 x^{3}-20 x^{2}-10 x+33=0$, by forming and solving a new cubic equation with roots $x_{1}+x_{2}, x_{1}+x_{3}, x_{2}+x_{3}$.
## Solution.

1. The required formulas are obtained by expanding the brackets of the given identity and combining like terms:

$$
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=\frac{c}{a}, x_{1} x_{2} x_{3}=-\frac{d}{a}
$$

These formulas are Vieta's theorem for cubic equations.

2. The new cubic equation is as follows:

$$
\left(x-\left(x_{1}+x_{2}\right)\right)\left(x-\left(x_{2}+x_{3}\right)\right)\left(x-\left(x_{1}+x_{3}\right)\right)=0
$$

By Vieta's theorem, $x_{1}+x_{2}+x_{3}=\frac{5}{2}$, so this equation can be written as: $\left(x-\frac{5}{2}+x_{1}\right)\left(x-\frac{5}{2}+x_{2}\right)\left(x-\frac{5}{2}+x_{3}\right)=0 \Leftrightarrow\left(\frac{5}{2}-x-x_{1}\right)\left(\frac{5}{2}-x-x_{2}\right)\left(\frac{5}{2}-x-x_{3}\right)=0$.

Let $y=\frac{5}{2}-x$, then $\left(y-x_{1}\right)\left(y-x_{2}\right)\left(y-x_{3}\right)=0 \Leftrightarrow$

$\Leftrightarrow 8 y^{3}-20 y^{2}-10 y+33=0 \Leftrightarrow 8\left(\frac{5}{2}-x\right)^{3}-20\left(\frac{5}{2}-x\right)^{2}-10\left(\frac{5}{2}-x\right)+33=0 \Leftrightarrow$

$\Leftrightarrow x^{3}-5 x^{2}+5 x-1=0 \Leftrightarrow\left(x^{3}-1\right)-5\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}-4 x+1\right)=0 \Leftrightarrow$ $\Leftrightarrow x=1$ or $x=2 \pm \sqrt{3}$. But the roots of the new equation are $\frac{5}{2}-x_{1}, \frac{5}{2}-x_{2}, \frac{5}{2}-x_{3}$. Therefore, $\frac{5}{2}-x_{1}=1, \frac{5}{2}-x_{2}=2+\sqrt{3}, \frac{5}{2}-x_{3}=2-\sqrt{3}$.

Answer: The roots of the equation are: $x_{1}=1.5 ; x_{2,3}=0.5 \pm \sqrt{3}$.