Solve each of the following systems of equations by the method of cross-multiplication :

GIVEN:

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

$x+y=2 a^{2}$

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

$x\left((a-b)+\frac{a b}{a-b}\right)-y\left((a+b)-\frac{a b}{a+b}\right)=0$

$x+y-2 a^{2}=0$

By cross multiplication method we get

$\frac{x}{\left(\left(-2 a^{2}\right) \times-\left((a+b)-\frac{a b}{a+b}\right)\right)-0}=\frac{-y}{\left(-2 a^{2}\right) \times\left((a-b)+\frac{a b}{a-b}\right)-0}$

$=\frac{1}{\left((a-b)+\frac{a b}{a-b}\right)-\left(-\left((a+b)-\frac{a b}{a+b}\right)\right)}$

$\frac{x}{\left(\left(-2 a^{2}\right) \times-\left(\frac{(a+b)^{2}-a b}{a+b}\right)\right)}=\frac{-y}{\left(-2 a^{2}\right) \times\left(\frac{(a-b)^{2}+a b}{a-b}\right)}$

$=\frac{1}{\left(\frac{(a-b)^{2}+a b}{a-b}\right)-\left(-\left(\frac{(a+b)^{2}-a b}{a+b}\right)\right)}$

$\frac{x}{\left(\left(-2 a^{2}\right) \times-\left(\frac{\left(a^{2}+b^{2}+2 a b\right)-a b}{a+b}\right)\right)}=\frac{-y}{\left(-2 a^{2}\right) \times\left(\frac{\left(a^{2}+b^{2}-2 a b\right)+a b}{a-b}\right)}$

$=\frac{1}{\left(\frac{\left(a^{2}+b^{2}-2 a b\right)+a b}{a-b}\right)-\left(-\left(\frac{\left(a^{2}+b^{2}+2 a b\right)-a b}{a+b}\right)\right)}$

$\frac{x}{\left(\left(-2 a^{2}\right) \times-\left(\frac{\left(a^{2}+b^{2}+a b\right)}{a+b}\right)\right)}=\frac{-y}{\left(-2 a^{2}\right) \times\left(\frac{\left(a^{2}+b^{2}-a b\right)}{a-b}\right)}$

$=\frac{1}{\left(\frac{\left(a^{2}+b^{2}-a b\right)}{a-b}\right)-\left(-\left(\frac{\left(a^{2}+b^{2}+a b\right)}{a+b}\right)\right)}$

$\frac{\frac{x}{\left(2 a^{4}+2 a^{2} b^{2}+2 a^{3} b\right)}}{a+b}=\frac{y}{\frac{\left(2 a^{4}+2 a^{2} b^{2}-2 a^{3} b\right)}{a-b}}$

$=\frac{1}{\left(\frac{\left(a^{2}+b^{2}-a b\right)}{a-b}\right)-\left(-\left(\frac{\left(a^{2}+b^{2}+a b\right)}{a+b}\right)\right)}$

$\frac{x}{\frac{\left(2 a^{4}+2 a^{2} b^{2}+2 a^{3} b\right)}{a+b}}=\frac{y}{\frac{\left(2 a^{4}+2 a^{2} b^{2}-2 a^{3} b\right)}{a-b}}$

$=\frac{1}{\left(\frac{\left(a^{2}+b^{2}-a b\right)(a+b)+\left(a^{2}+b^{2}+a b\right)(a-b)}{(a-b)(a+b)}\right)}$

$\frac{x}{\frac{\left(2 a^{4}+2 a^{2} b^{2}+2 a^{3} b\right)}{a+b}}=\frac{\frac{1}{\left(2 a^{4}+2 a^{2} b^{2}-2 a^{3} b\right)}}{a-b}=\frac{1}{\left(\frac{\left(a^{3}+b^{3}+a^{3}-b^{3}\right)}{(a-b)(a+b)}\right)}$

$\frac{x}{\frac{\left(2 a^{4}+2 a^{2} b^{2}+2 a^{3} b\right)}{a+b}}=\frac{y}{\frac{\left(2 a^{4}+2 a^{2} b^{2}-2 a^{3} b\right)}{a-b}}=\frac{1}{\left(\frac{2 a^{3}}{(a-b)(a+b)}\right)}$

Consider the following for x

$\frac{x}{\frac{\left(2 a^{4}+2 a^{2} b^{2}+2 a^{3} b\right)}{a+b}}=\frac{1}{\left(\frac{2 a^{3}}{(a-b)(a+b)}\right)}$

$\frac{x}{\frac{\left(a^{2}+b^{2}+a b\right)}{a+b}}=\frac{1}{\left(\frac{a}{(a-b)(a+b)}\right)}$

$x\left(\frac{a}{(a-b)(a+b)}\right)=\frac{\left(a^{2}+b^{2}+a b\right)}{a+b}$

$x=\frac{\left(a^{2}+b^{2}+a b\right)(a-b)}{a}$

$x=\frac{\left(a^{3}+a b^{2}+a^{2} b-b^{3}-a b^{2}-a^{2} b\right)}{a}$

$x=\frac{\left(a^{3}-b^{3}\right)}{a}$

And 

$\frac{y}{\frac{\left(2 a^{4}+2 a^{2} b^{2}-2 a^{3} b\right)}{a-b}}=\frac{1}{\left(\frac{2 a^{3}}{(a-b)(a+b)}\right)}$

$\frac{y}{\frac{\left(a^{2}+b^{2}-a b\right)}{a-b}}=\frac{1}{\left(\frac{a}{(a-b)(a+b)}\right)}$

$\frac{y}{\frac{\left(a^{2}+b^{2}-a b\right)}{a-b}}=\frac{1}{\left(\frac{a}{(a-b)(a+b)}\right)}$

$y\left(\frac{a}{(a-b)(a+b)}\right)=\frac{\left(a^{2}+b^{2}-a b\right)}{a-b}$

$y=\frac{\left(a^{2}+b^{2}-a b\right)(a+b)}{a}$

$y=\frac{\left(a^{3}+b^{3}\right)}{a}$

Hence we get the value of $x=\frac{a^{3}-b^{3}}{a}$ and $y=\frac{a^{3}+b^{3}}{a}$