Factorise the following, using the identity, $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ :

$9 x^{2} + 2 x y + \frac{y^{2}}{9}$

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Solution

We have, $9 x^{2} + 2 x y + \frac{y^{2}}{9}$
$= (3 x)^{2} + 2 \times 3 x \times \frac{y}{3} + (\frac{y}{3})^{2} = (3 x + \frac{y}{3})^{2} = (3 x + \frac{y}{3}) (3 x + \frac{y}{3})$

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