$${ \left(3a- \frac{ 3 }{ 4 } b \right) }^{ 2 }$$
$\frac{9\left(4a-b\right)^{2}}{16}$
$$9a^{2}-\frac{9}{2}ab+\frac{9}{16}b^{2}$$
Show Solution
Hide Solution
$-\frac{9ab}{2}+\frac{9b^{2}}{16}+9a^{2}$
