Equations with unknowns and powers $a + b +c = 17$
$a^2 + b^2 + c^2 = 101$
$a^3 + b^3 + c^3 = 623$
How does one go about solving this?
Thanks
 A: Hint: 
Expand $(a + b + c)^2$ and $(a + b + c)^3$, and use the three equations you are given.
$$(a+b+c)^2 = (17)^2 = \underbrace{\color{blue}{a^2 + b^2 + c^2}}_{101} +2(\bf{ab + ac + bc})$$
$$\begin{align} (a + b+ c)^3 = (17)^3 & = \underbrace{\color{red}{a^3 + b^3 + c^3}}_{623} + 3(a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2) \\ \\&= 623 + 3(a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2 + \color{green}{abc - abc})\\ \\ & = 623 + 3\Big(\underbrace{(a + b + c)}_{17}{\bf(ab + ac + bc)}- \color{green}{\bf abc}\Big)\end{align}$$  
Spoiler: 

 Try the triple of numbers $4, 6, 7$.

A: my weaknesses show up particularly when doing tedious calculations, so as a penance, just this once, i will bludgeon my way through this, probably generating more heat than light in the process!
we have: $$a+b=17-c$$
thus
$$(a+b)^2 = 289 -34c+c^2$$
and 
$$(a+b)^3 = 4913 - 867c+51c^2 -c^3$$
but $$(a+b)^2 = a^2+b^2 + 2ab = 101 - c^2 +2ab$$
giving
$$ab = 94 -17c +c^2$$
likewise $$(a+b)^3 = a^3+b^3 +3ab(a+b) \\
= 623 -c^3 +3(94-17c+c^2)(17-c) \\
=5417 -1149c +102c^2 -4c^3$$
so we have 
$$ 5417 -1149c +102c^2 -4c^3 = = 4913 -867c+51c^2 -c^3$$
i.e.
$$504 -282c+51c^2 -3c^3 =0$$
or
$$c^3-17c^2+94c-168 =0$$
or
$$(c-4)(c-6)(c-7) = 0$$
