# Question #70432

Mar 24, 2017

See below.

#### Explanation:

$\left(x + {x}_{1}\right) \left(x + {x}_{2}\right) \left(x - {x}_{3}\right) = {x}^{3} + \left({x}_{1} + {x}_{2} + {x}_{3}\right) {x}^{2} + \left({x}_{1} {x}_{2} + {x}_{1} {x}_{3} + {x}_{2} {x}_{3}\right) x + {x}_{1} {x}_{2} {x}_{3}$ so

$\left\{\begin{matrix}{x}_{1} + {x}_{2} + {x}_{3} = a \\ {x}_{1} {x}_{2} + {x}_{1} {x}_{3} + {x}_{2} {x}_{3} = b \\ {x}_{1} {x}_{2} {x}_{3} = c\end{matrix}\right.$

We are looking for a polynomial ${x}^{3} + \alpha {x}^{2} + \beta x + \gamma$ such that

$\left\{\begin{matrix}{x}_{1}^{3} + {x}_{2}^{3} + {x}_{3}^{3} = \alpha \\ {x}_{1}^{3} {x}_{2}^{3} + {x}_{1}^{3} {x}_{3}^{3} + {x}_{2}^{3} {x}_{3}^{3} = \beta \\ {x}_{1}^{3} {x}_{2}^{3} {x}_{3}^{3} = \gamma\end{matrix}\right.$

Clearly we have

$\gamma = {c}^{3}$

now ${\left({x}_{1} + {x}_{2} + {x}_{3}\right)}^{3} = {x}_{1}^{3} + {x}_{2}^{3} + {x}_{3}^{3} + 3 \left({x}_{1}^{2} \left({x}_{2} + {x}_{3}\right) + {x}_{2}^{2} \left({x}_{1} + {x}_{3}\right) + {x}_{3}^{2} \left({x}_{1} + {x}_{2}\right)\right) + 6 {x}_{1} {x}_{2} {x}_{3}$

but

${x}_{1}^{2} \left({x}_{2} + {x}_{3}\right) + {x}_{2}^{2} \left({x}_{1} + {x}_{3}\right) + {x}_{3}^{2} \left({x}_{1} + {x}_{2}\right) = \left({x}_{1} + {x}_{2} + {x}_{3}\right) \left({x}_{1} {x}_{2} + {x}_{1} {x}_{3} + {x}_{2} {x}_{3}\right) + 3 {x}_{1} {x}_{2} {x}_{3}$

so we have

${a}^{3} = {x}_{1}^{3} + {x}_{2}^{3} + {x}_{3}^{3} + 3 \left(a b + 3 c\right) + 6 c$

and then

$\alpha = {a}^{3} - 3 \left(a b + 3 c\right) - 6 c$

now making

${\left({x}_{1} {x}_{2} + {x}_{1} {x}_{3} + {x}_{2} {x}_{3}\right)}^{3} = {x}_{1}^{3} {x}_{2}^{3} + {x}_{1}^{3} {x}_{3}^{3} + {x}_{2}^{3} {x}_{3}^{3} + 3 \left({x}_{1}^{3} {x}_{2}^{2} {x}_{3} + {x}_{1}^{2} {x}_{2}^{3} {x}_{3} + {x}_{1}^{3} {x}_{2} {x}_{3}^{2} + {x}_{1} {x}_{2}^{3} {x}_{3}^{2} + {x}_{1}^{2} {x}_{2} {x}_{3}^{3} + {x}_{1} {x}_{2}^{2} {x}_{3}^{3}\right) + 6 {x}_{1}^{2} {x}_{2}^{2} {x}_{3}^{2}$

but

${x}_{1}^{3} {x}_{2}^{2} {x}_{3} + {x}_{1}^{2} {x}_{2}^{3} {x}_{3} + {x}_{1}^{3} {x}_{2} {x}_{3}^{2} + {x}_{1} {x}_{2}^{3} {x}_{3}^{2} + {x}_{1}^{2} {x}_{2} {x}_{3}^{3} + {x}_{1} {x}_{2}^{2} {x}_{3}^{3} = {x}_{1} {x}_{2} {x}_{2} \left({x}_{1}^{2} \left({x}_{2} + {x}_{3}\right) + {x}_{2}^{2} \left({x}_{1} + {x}_{3}\right) + {x}_{3}^{2} \left({x}_{1} + {x}_{2}\right)\right)$

so finally

${b}^{3} = \beta + 9 c \left(a b + 3 c\right) + 6 {c}^{3}$

and then

$\beta = {b}^{3} - 9 c \left(a b + 3 c\right) - 6 {c}^{3}$ so the sought polynomial is

${x}^{3} + \left({a}^{3} - 3 \left(a b + 3 c\right) - 6 c\right) {x}^{2} + \left({b}^{3} - 9 c \left(a b + 3 c\right) - 6 {c}^{3}\right) x + {c}^{3}$