Find $k$ such that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$ holds for the roots of $x^3-6x^2+kx+k=0$

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?

• at first i would expand $$(\alpha-1)^3+(\beta-2)^3+\gamma-3)^3=0$$ so working backwards – Dr. Sonnhard Graubner Jul 24 '17 at 20:01

Since $\alpha-1+\beta-2+\gamma-3=0$ and $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we see that $$(\alpha-1)(\beta-2)(\gamma-3)=0$$ and we have three following cases:
1. $\alpha=1$, which gives $k=\frac{5}{2}$ and our equation is $$2x^3-12x^2+5x+5=0$$ or $$2x^3-2x^2-10x^2+10x-5x+5=0$$ or $$(x-1)(2x^2-10x-5)=0,$$ which gives $\left\{1,\frac{5+\sqrt{35}}{2},\frac{5+\sqrt{35}}{2}\right\}$.