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

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
$$(c-4)(c-6)(c-7) = 0$$
