If ac a c = cd c d = k, find the value of 3a2?ac+c23b2?bd+d2

If $\frac{a}{c}$ = $\frac{c}{d}$ = k, find the value of $\frac{3a^2?ac+c^2}{3b^2?bd+d^2}$

Answer Details

$\frac{a}{c}$ = $\frac{c}{d}$ = k
∴ $\frac{a}{b}$ = bk
$\frac{c}{d}$ = k
∴ c = dk
= $\frac{3a^2-ac+c^2}{3b^2-bd+d^2}$
= $\frac{3(bk{)}^2-(bk\left)\right(dk)+d{k}^2}{3b^2-bd+a^2}$
= $\frac{3b^2{k}^2-b{k}^{2}d+d{k}^2}{3b^2-bd+d^2}$
k = $\frac{3b^2{k}^2-b{k}^{2}d+d{k}^2}{3b^2-bd+d^2}$