$$\frac { 1 } { 16 } x ^ { 2 } - \frac { 1 } { 12 } x y + \frac { 1 } { 36 } y ^ { 2 }$$
$\frac{\left(3x-2y\right)^{2}}{144}$
$$\frac{9x^{2}-12xy+4y^{2}}{144}$$
$$\left(3x-2y\right)^{2}$$
$$\frac{\left(3x-2y\right)^{2}}{144}$$
Show Solution
Hide Solution
