What is the value of $\frac{4x^2+9y^2+12xy}{144}$ ?

$(\frac{x}{3}+\frac{y}{4})^2$ $(\frac{x}{3}+y)^2$ $(\frac{x}{4}+\frac{y}{6})^2$ $(\frac{x}{6}+\frac{y}{4})^2$

$(\frac{x}{6}+\frac{y}{4})^2$

$\frac{4x^2+9y^2+12xy}{144}$ ( a + b ) = \(\frac{a^3 + b^3}{a^2 + b^2 - ab }\) = $\frac{(2x+3y)^2}{(12)^2}$ = [\(\frac{2x}{12}\) +  \(\frac{3x}{12}\) ]2 = $(\frac{x}{6}+\frac{y}{4})^2$