6.252. Find the coefficients of the equation $x^{2}+p x+q=0$ given that the difference between the roots of the equation is 5, and the difference between their cubes is 35.
## Solution.

From the condition we have

$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{1}-x_{2}=5, \\
x_{1}^{3}-x_{2}^{3}=35
\end{array} \Leftrightarrow\left\{\begin{array}{l}
x_{1}-x_{2}=5, \\
\left(x_{1}-x_{2}\right.
\end{array}\right)\left(\left(x_{1}-x_{2}\right)^{2}+3 x_{1} x_{2}\right)=35\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x _ { 1 } - x _ { 2 } = 5 , } \\
{ 5 ( 2 5 + 3 x _ { 1 } x _ { 2 } ) = 3 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x _ { 1 } - x _ { 2 } = 5 , } \\
{ x _ { 1 } x _ { 2 } = - 6 , }
\end{array} \text { from which } \left\{\begin{array} { l } 
{ x _ { 1 } = 2 , } \\
{ x _ { 2 } = - 3 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=3, \\
x_{2}=-2
\end{array}\right.\right.\right.\right.
\end{aligned}
$$

By Vieta's theorem we have:

1) $\left\{\begin{array}{l}p_{1}=-\left(x_{1}+x_{2}\right)=1 \\ q_{1}=x_{1} \cdot x_{2}=-6\end{array}\right.$
2) $\left\{\begin{array}{l}p_{2}=-1 \\ q_{2}=-6\end{array}\right.$

Answer: $p_{1}=1, q_{1}=-6 ; p_{2}=-1, q_{2}=-6$.