\(ab^2+a^2b=6\Leftrightarrow ab\left(b+a\right)=6\Leftrightarrow ab=2\)

So,a,b are 2 roots of the equation : \(x^2-3x+2=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(\Rightarrow\left(a;b\right)=\left(1;2\right);\left(2;1\right)\Rightarrow a^5+b^5=1^5+2^5=33\)

The solution of (a;b) is (1;2) and (2;1) => These two solutions are equal.

When (a;b) equal (1;2) => \(a^5+b^5=1^5+2^5=33\)

When (a;b) equal (2;1) => \(a^5+b^5=2^5+1^5=33\)