Wanna get boxed?

$\frac { { a }^{ 4 }+{ a }^{ 2 }{ b }^{ 2 }+{ b }^{ 4 } }{ { a }^{ 2 }+ab+{ b }^{ 2 } } ={ a }^{ 2 }-ab+{ b }^{ 2 }$ $\frac { 133 }{ 19 } ={ a }^{ 2 }-ab+{ b }^{ 2 }$ ${ a }^{ 2 }-ab+{ b }^{ 2 }=7$ ${ a }^{ 2 }+ab+{ b }^{ 2 }-({ a }^{ 2 }-ab+{ b }^{ 2 })=19-7$ $2ab=12$ $ab=6$

squaring the first eqn $a^{4} + a^{2}b^{2} + b^{4} + 2ab(a^{2} + b^{2} +ab) = 361$ $133 + 2ab*19 = 361$ hence $ab = 6$

$\begin{aligned}a^4+a^2b^2+b^4=&\ (a^4+2a^2b^2+b^4)-a^2b^2\\133=&\ (a^2+b^2)^2-(ab)^2\\133=&\ (a^2+b^2+ab)(a^2+b^2-ab)\\133=&\ 19[(a^2+ab+b^2)-2ab]\\7=&\ 19-2ab\\-12=&\ -2ab\\ab=&\ \boxed6\end{aligned}$