Finding the implicit constraint in a question

Question

I came across this question.

$$ a+b+c=11 \\ ab+bc+ca=25 \\ a^3+b^3+c^3+3abc=? $$

While trying to solve this I ended up with $abc=0$ and further solving gave $a= \frac{11-\sqrt{21}}{2}$, $b= \frac{11+\sqrt{21}}{2}$ and $c=0$. But, I am unable to see how two equations in the three unknowns yields an unique solution. Is there an implicit constraint or are there other solutions I am failing to see?

EDIT: My line of solution was as follows -

$$ \begin{align*} a^2+b^2+c^2 &= (a+b+c)^2 - 2(ab+bc+ca) \\ &= 11*11 - 50 = 71 \\ (a+b+c)^3 &= a^3 + b^3 + c^3 + 3a(b^2+c^2)+3b(a^2+c^2)+3c(b^2+c^2) +3abs \\ &= a^3 + b^3 + c^3 + 3a(71 - a^2) + 3b(71 - b^2) + 3c(71 - c^2) + 3abc\\ 11^3 &= -2 (a^3 + b^3 + c^3) + 3*71*(a+b+c) +3abc \\ 2(a^3 + b^3 + c^3) - 3abc &= (71*3 - 11^2)*11 = 92*11 \end{align*} $$

I then arrived at mcd's comment like this

$$ \begin{align*} (a^2+b^2+c^2)(a+b+c)&=a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)\\ 71*11&=a^3+b^3+c^3-3abc+11*25 \\ a^3+b^3+c^3-3abc&=46*11 \end{align*} $$

Solving the two simultaneous equations in $a^3+b^3+c^3$ and $abc$, I got $abc=0$ and rest followed. But I feel I am making a mistake some where that is making me lose other solutions to the given two equations.

Answer 1

One way is to show that $b$ and $c$ can be expressed by $a$ with a quadratic equation. Let $d=- 3a^2 + 22a + 21$, then we either have $$ b=\frac{11-a+\sqrt{d}}{2},\; c=\frac{11-a-\sqrt{d}}{2}, $$ or $b$ and $c$ interchanged. So these $(a,b,c)$ with arbitrary $a$ are the solutions.

In both cases, of course, then the third equation has the same value, which is $$ 2(3a^3 - 33a^2 + 75a + 253). $$ However, this value will depend on $a$. For example, with $a=0$, we obtain $b$ and $c$ as in the post, $$ b=\frac{11+\sqrt{21}}{2},\; c=\frac{11-\sqrt{21}}{2} $$ and the third equation has the value $506$.

Answer 2

COMMENT(very pertinent).

One has $$(a+b+c)^3=a^3+b^3+c^3+6abc+3[+a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)]\\(a+b+c)^3=a^3+b^3+c^3+6abc+3[ab(11-c)+ac(11-b)+bc(11-a)]\\(a+b+c)^3=a^3+b^3+c^3+6abc+3[11(ab+ac+bc)-3abc]\\(a+b+c)^3=a^3+b^3+c^3-3abc+3\cdot11\cdot25$$ Finally we have $$a^3+b^3+c^3-3abc=11^3-3\cdot11\cdot25=506$$ and this value is not dependent of a free choice of $a$.

But also has been shown above, choosing $a=0$, that $$a^3+b^3+c^3+3abc=506$$

The two first equations form a system which have no integer solutions but a quadratic one with a parameter (it could be any of $a,b,c$).

What I want to highlight is that Discretizer said in his post about $abc=0$ seems not to be a mistake as he believed. The question remains when $a$ is not choosen equal to zero in Dietrich Burde's answer.

Answer 3

$a^3 + b^3 + c^3 - 3abc$ really is constant, value 506.

But $a^3 + b^3 + c^3 + 3abc$ is not constant.

Note: the equations say $a^2 + b^2 + c^2 = 71,$ and the intersection of this sphere with the plane $a+b+c = 11 $ is a circle.

I took $$ a = \frac{11 + \sqrt{138}}{3} \; , \; \; b = \frac{11 - \sqrt{138}}{3} \; , \; \; c = \frac{11 }{3} \; . \; \; $$ Then $$a^3 + b^3 + c^3 + 3abc= \frac{4180}{9} \approx 464.44$$

The two given equations say that the collection of $(a,b,c)$ points lie in a circle with center at $(11,11,11) / 3 $ and a certain radius, which I kept getting wrong. I started over more simply, knowing that there was a point of the circle to be found by taking the multiple of $(1, -1,0)$ from the center $(11,11,11) / 3 ,$ because $(1, -1,0)$ and $(1, 1,1)$ are perpendicular. That worked. That's how I picked the $\sqrt{138}$ stuff.

Answer 4

$x^3-e_1x^2+e_2x-e_3=0$ $e_1=a+b+c=11$ $e_2=ab+bc+ca=25\\ e_3=abc$

By Newton-Girard identities, $p_3=e_1^3-3e_1e_2+3e_3$

Hence, $p_3-3e_3=e_1(e_1^2-3e_2)=11\times46=506$

I don't think that $p_3+3e_3$ is a constant.