# Question about roots

Let $a,b,c$ be roots of equation $x^3-6x^2+kx+k=0$,and $(a-1)^3+(b-2)^3+(c-3)^3=0$.

how to compute $a,b,c,k=?$

if we do work equivalently as to find out the solution: \begin{align*}(1)&abc=&-k\\(2)&ab+ac+bc=&k\\(3)&a+b+c=&6\\(4)&(a-1)^3+(b-2)^3+(c-3)^3=&0\end{align*}

Is there good skill?

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Interesting question (+1) – user 1618033 Feb 7 '13 at 11:01

We have $a+b+c=6, ab+ac+bc=k, abc=-k, (a-1)^3+(b-1)^3+(c-1)^3=0$.

Thus $(a+1)(b+1)(c+1)=abc+(ab+ac+bc)+(a+b+c)+1=k+(-k)+6+1=7$.

Substitute $x=a-1, y=b-2, z=c-3$, so $x+y+z=x^3+y^3+z^3=0, \, (x+2)(y+3)(z+4)=7$.

Note that conversely any $x, y, z$ satisfying these equations will in turn give unique $a, b, c$ satisfying $a+b+c=6, (a-1)^3+(b-2)^3+(c-3)^3=0, (a+1)(b+1)(c+1)=7$, so that $abc+ab+ac+bc=(a+1)(b+1)(c+1)-(a+b+c+1)=7-6-1=0$. Now from the values of $a, b, c$, we can then uniquely determine $k$.

Now $x^3+y^3=-z^3=(-z)^3=(x+y)^3$, so $0=3xy(x+y)=-3xyz$.

Thus at least 1 of $x, y, z$ is 0.

If $x=0$, then $z=-y$, so $2(y+3)(4-y)=7$, so $2y^2-2y-17=0$, giving $y=\frac{1 \pm \sqrt{35}}{2}$.

We have $(x, y, z)=(0, \frac{1 \pm \sqrt{35}}{2}, -\frac{1 \pm \sqrt{35}}{2})$.

If $y=0$, then $z=-x$, so $(x+2)(3)(4-x)=7$, so $3x^2-6x-17=0$, giving $x=\frac{3 \pm \sqrt{60}}{3}$.

We have $(x, y, z)=(\frac{3 \pm \sqrt{60}}{3}, 0, -\frac{3 \pm \sqrt{60}}{3})$.

If $z=0$, then $y=-x$, so $(x+2)(3-x)(4)=7$, so $4x^2-4x-17=0$, giving $x=\frac{1 \pm \sqrt{18}}{2}$.

We have $(x, y, z)=(\frac{1 \pm \sqrt{18}}{2}, -\frac{1 \pm \sqrt{18}}{2}, 0)$.

Thus $(a, b, c)=(1, 2+\frac{1 \pm \sqrt{35}}{2}, 3-\frac{1 \pm \sqrt{35}}{2}), (1+\frac{3 \pm \sqrt{60}}{3}, 2, 3-\frac{3 \pm \sqrt{60}}{3}), (1+\frac{1 \pm \sqrt{18}}{2}, 2-\frac{1 \pm \sqrt{18}}{2}, 3)$.

Edit: Calculating values of $k$ using $k=-abc$ now gives $(a, b, c, k)=(1, 2+\frac{1 \pm \sqrt{35}}{2}, 3-\frac{1 \pm \sqrt{35}}{2}, \frac{5}{2}), (1+\frac{3 \pm \sqrt{60}}{3}, 2, 3-\frac{3 \pm \sqrt{60}}{3}, \frac{16}{3}), (1+\frac{1 \pm \sqrt{18}}{2}, 2-\frac{1 \pm \sqrt{18}}{2}, 3, \frac{27}{4})$.

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Hints:

\;\;\begin{align*}(1)&abc=&-k\\(2)&ab+ac+bc=&k\\(3)&a+b+c=&6\end{align*}

You may want to read about Viéte's formulae , too.

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I know this theorem. but also hard to find out the solution. – Leitingok Feb 7 '13 at 11:31
Did anyone promise you easy work in mathematics? Because if someone did then he/she/they lied to you. At least here you have an idea to begin working, so downvoting the answer seems a little too harsh, doesn't it? – DonAntonio Feb 7 '13 at 11:51
I mean this meathod is not easy without using software such as matlab. – Leitingok Feb 7 '13 at 11:53
Yes... so? Spend some time making some calculations, what's the big deal?! For example $$a^3+b^3+c^3=\left(a+(b+c)\right)^3-3a^2(b+c)-3a(b+c)^2-3b^2c-3bc^2=$$ $$=\left(a+b+c\right)^3-3\left(a(ab+ab+bc)+b(ab+ac+bc)\right)\ldots\;\;etc.$$ – DonAntonio Feb 7 '13 at 12:05
not this.I did some work that you give me. but it's not easy to use in $(a-1)^3+(b-2)^3+(c-3)^3$ – Leitingok Feb 7 '13 at 12:12