Factorise the following using appropriate identities: 

(i) $9x^2+6xy+y^2$ (ii) $4y^2-4y+1$ (iii) $x^2-\frac{y^2}{100}$

(i) Using the algebraic identity: ${\left(x+y\right)}^2=x^2+2xy+y^2$

We have: $x=3x; y=y$

$\begin{array}{rcl}9x^2+6xy+y^2& =& {\left(3x\right)}^2+2\left(3x\right)\left(y\right)+{\left(y\right)}^2\\ & =& {\left(3x+y\right)}^2\\ & =& \left(3x+y\right)\left(3x+y\right)\end{array}$

(ii) Using the algebraic identity: ${\left(x-y\right)}^2=x^2-2xy+y^2$

We have: $x=2y; y=1$

$\begin{array}{rcl}4y^2-4y+1& =& {\left(2y\right)}^2-2\left(2y\right)\left(1\right)+{\left(1\right)}^2\\ & =& {\left(2y-1\right)}^2\\ & =& \left(2y-1\right)\left(2y-1\right)\end{array}$

(iii) Using the algebraic identity: $x^2-y^2=\left(x+y\right)\left(x-y\right)$

We have: $x=x; y=\frac{y}{10}$

$\begin{array}{rcl}{x}^2-\frac{y^2}{100}& =& {\left(x\right)}^2-{\left(\frac{y}{10}\right)}^2\\ & =& \left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)\end{array}$