Factorise the following using appropriate identities: <br/><br/>(i) $9 x^{2}+6 x y+y^{2}$<br/><br/> (ii) $4 y^{2}-4 y+1$ <br/><br/>(iii) $x^{2}-\frac{y^{2}}{100}$

Solution:

(i) $9 x^{2}+6 x y+y^{2}=(3 x)^{2}+2(3 x)(y)+(y)^{2}$

$=(3 x+y)(3 x+y)$

$\left[x^{2}+2 x y+y^{2}=(x+y)^{2}\right]$

(ii) $4 y^{2}-4 y+1=(2 y)^{2}-2(2 y)(1)+(1)^{2}$

$=(2 y-1)(2 y-1)$

$\left[x^{2}-2 x y+y^{2}=(x-y)^{2}\right]$

(iii) $x^{2}-\frac{y^{2}}{100}=x^{2}-\left(\frac{y}{10}\right)^{2}$

$=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)$

$\left[x^{2}-y^{2}=(x+y)(x-y)\right]$