Come and See this Nice Real System

Step 1) Use the following identity for first equation $x^3+y^3+z^3-3xyz=(x+y+z) \left( \frac{(x-y)^2+(x-z)^2+(y-z)^2}{2} \right)$ So we can write $a^3+6ab+8-b^3=a^3+(-b)^3+2^3-3(a)(-b)(2)=0 \\ (a-b+2)((a+b)^2+(a-2)^2+(b+2)^2)=0$ Therefore we have $b=a+2$ or $a=-b=2$

Step 2) Using the obtained possibilities for $a$ and $b$ in the second equation. Second case i.e. $a=-b=2$ does not satisfy second equation. One possibility remains and that is $b=a+2$ which gives us $a^5+32=(a+2)^5 \\ a^4+4a^3+8a^2+8a=0 \\ a(a+2)(a^2+2a+4)=0 \\ a=0,-2$ For $a=0$ we get $b=2$ and for $a=2$ we get $b=0$ . Finally there are two such pairs $(a,b)=(0,2), (-2,0)$ Answer will be $2+(0)^2+(2)^2+(-2)^2+(0)^2=10$

My idea was identical to the Kazem. This identity solve several interesting problems.

Using the cubic formula, we solve the first equation for $b$ and find $b=a+2$ . (We don't have to worry about the special case $a=2$ , when the discriminant is 0, since the system has no solutions in that case.) The system now reduces to $(a+2)^5=a^5+32$ ; consider the function $f(a)=(a+2)^5-a^5-32$ . We can spot the roots $f(0)=f(-2)=0$ . There can't be any other solutions since the graph of $f(a)$ is concave up; indeed, $f''(a)=40(3a^2+6a+4)>0$ for all $a.$

Thus the only solutions are $(0,2)$ and $(-2,0).$