> Austria 1983
Find all $a \in \mathbb{R}$ such that the roots $\alpha, \beta, \gamma$ of
$
x^3-6 x^2+a x+a=0
$
satisfy
$
(\alpha-1)^3+(\beta-1)^3+(\gamma-1)^3=0
$
> [!sol]- Click for Solution!
> Write the polynomial factored to see how the roots relate to the coefficients (Vieta's)
> $\begin{align}
x^3-6 x^2+a x+a&=(x-\alpha)(x-\beta)(x-\gamma)\\
&=x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\alpha \gamma) x-\alpha \beta \gamma
\end{align}$
> Equating the coefficients gives 3 equations.
> $\begin{aligned} \alpha+\beta+\gamma&=6 \\ \alpha \beta+\beta \gamma+\alpha \gamma&=a \\ \alpha \beta \gamma&=-a\end{aligned}$
> The nature of the roots $\alpha,$ $\beta,$ $\gamma$ allows one more equation to be made.
> $\begin{aligned} \alpha^3-6 \alpha^2+a \alpha+a&=0 \\ \beta^3-6 \beta^2+a \beta+a&=0 \\ \gamma^3-6 \gamma^2+a \gamma+a&=0\\\hline\left(\alpha^3+\beta^3+\gamma^3\right)-6\left(\alpha^2+\beta^2+\gamma^2\right)+a\left(\alpha+\beta+\gamma\right)+3a&=0\end{aligned}$
> Now we introduce the desired condition, expand it, and group it by degrees.
> $\begin{align}
(\alpha-1)^3+(\beta-1)^3+(\gamma-1)^3&=0\\
\left(\alpha^3-3 \alpha^2+3 \alpha-1\right)+\left(\beta^3-3 \beta^2+3 \beta-1\right)+\left(\gamma^3-3 \gamma^2+3 \gamma-1\right)&=0 \\ \left(\alpha^3+\beta^3+\gamma^3\right)-3\left(\alpha^2+\beta^2+\gamma^2\right)+3(\alpha+\beta+\gamma)-3&=0
\end{align}$
> Subtract a previous equation
> $\begin{align}\left(\alpha^3+\beta^3+\gamma^3\right)-3\left(\alpha^2+\beta^2+\gamma^2\right)+3(\alpha+\beta+\gamma)-3&=0\\\left(\alpha^3+\beta^3+\gamma^3\right)-6\left(\alpha^2+\beta^2+\gamma^2\right)+a\left(\alpha+\beta+\gamma\right)+3a&=0\\\hline3\left(\alpha^2+\beta^2+\gamma^2\right)+(3-a)(\alpha+\beta+\gamma)-(3 a+3)&=0\end{align}$
> Recognize from earlier $\alpha+\beta+\gamma=6$
> $3\left(\alpha^2+\beta^2+\gamma^2\right)+(3-a)(6)-(3 a+3)=0$
> And simplify:
> $\begin{align}\left(\alpha^2+\beta^2+\gamma^2\right)+(3-a)(2)-(a+1)&=0\\\left(\alpha^2+\beta^2+\gamma^2\right)+5-3a&=0\\\left(\alpha^2+\beta^2+\gamma^2\right)&=3a-5\end{align}$
> Take a moment to realize $\left(\alpha^2+\beta^2+\gamma^2\right)$ is within reach, since we already have what we need to write it in terms of $a$:
> $\begin{align}\left(\alpha^2+\beta^2+\gamma^2\right)+2(\alpha\beta+\beta \gamma+\alpha \gamma)&=\left(\alpha+\beta+\gamma\right)^2\\\left(\alpha^2+\beta^2+\gamma^2\right)+2(a)&=\left(6\right)^2\\\left(\alpha^2+\beta^2+\gamma^2\right)&=36-2a\end{align}$
> Both expressions for $\left(\alpha^2+\beta^2+\gamma^2\right)$ should be equivalent:
> $\begin{align}36-2a &= 3a-5 \\ 41&=5a\\ \boxed{\frac{41}5}&=a\\\end{align}$