if $a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$ 
If $a+b$ is an positive integer and $a\ge b$ and
  $a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.

I tried a lot to solve it, i came to a step after which i was not able to proceed forward, the step was $(a+b)^3 - 41\cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$
Please help me to solve this.

Note: I solve it base on that a,b are positive numbers(not integers)
suppose $s=a+b$
Then, $a^3+(s-a)^3 +3(a^2+(s-a)^2) - 700s^2 = 0$
Rearrange it, $(6+3s)a^2-(3s^2+6s)a+(s^3-697s^2)=0$
$\Delta=(3s^2+6s)^2-4(6+3s)(s^3-697s^2)=-3s^4+8376s^3+16764s^2>0$
$0<s<=2794$
for one solution, $0<a<s$
(1)  $0<3s^2+6s-\sqrt\Delta<2s(6+3s)$
L:$3s^2+6s>\sqrt\Delta$ 
$(3s^2+6s)^2>(3s^2+6s)^2-4(6+3s)(s^3-697s^2)$
$4(6+3s)(s^3-697s^2)>0$
$0<s<697$
R:$-3s^2-6s<\sqrt\Delta$ It is always true
(2) $0<3s^2+6s+\sqrt\Delta<2s(6+3s)$
L: always true
R: $3s^2+6s>\sqrt\Delta$ 
It is true for $0<s<697$
Overall $a+b=s=1,2,...695$ or $696$
$sum=242556$
