If $\alpha$, $\beta$ and $\gamma$ are the three roots of the equation $x^3-6x^2+kx+k=0$, find the values of $k$ such that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$. For each of the possible values of $k$ solve the equation.
I conclude from the equation that $\alpha+\beta+\gamma=6$, $\alpha\beta+\beta\gamma+\gamma\alpha=k$ and $\alpha\beta\gamma=-k$ and with the condition that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$. I found the solution from wolfram: http://www.wolframalpha.com/input/?i=Solve%5Ba%2Bb%2Bc%3D6,ab%2Bbc%2Bca%3Dk,abc%3D-k,(a-1)%5E3%2B(b-2)%5E3%2B(c-3)%5E3%3D0%5D
From the solution I can see that either $\alpha=1$ or $\beta=2$ or $\gamma=3$. I guess to calculate the term $(\alpha-1)(\beta-2)(\gamma-3)$ and prove it to be zero. Should I continue with this or are there another other method?