Solve : `a(x+y)+b(x-y)=a^2-a b+b^2` `a(x+y)-b(x-y)=a^2+a b+b^2`

Explanation

Given equations can be written as
`ax + ay + bx - by = a^(2) - ab + b^(2)`
`implies (a+b) x + (a - b) y = a^(2) - ab + b^(2)" ....(1)"`
and `ax + ay - bx + by = a^(2) + ab + b^(2)`
`implies (a-b) x + (a+b) y = a^(2) + ab + b^(2) " ....(2)"`
For equations (1) and (2) by cross multiplication method, we have

`(x)/((a-b) xx [-a^(2) + ab + b^(2)]-(a+b) xx [-(a^(2) - ab + b^(2))])`
`=(y)/([-(a^(2)-ab+b^(2))]xx(a-b)-[-(a^(2)+ab + b^(2))xx (a + b)])=(1)/((a+b)(a+b)-(a-b)(a-b))`
implies `(x)/(-(a^(3) - b^(3))+ (a^(3)+b^(3)))`
`=(y)/(-(a^(3) - a^(2)b + ab^(2) - a^(2)b + ab^(2) - b^(3))+(a^(3)+a^(2)b+ab^(2)+a^(2)b+ab^(2)+b^(3)))=(1)/((a+b)^(2)-(a-b)^(2))`
implies `(x)/(2b^(3)) = (y)/(2a^(2)b-2ab^(2) + b^(3) + 2a^(2)b + 2ab^(2) + b^(3))=(1)/(4ab)`
implies `(x)/(2b^(3)) = (y)/(4a^(2)b + 2b^(3)) = (1)/(4ab)`
implies `(x)/(2b^(3))=(y)/(2b(2a^(2)+b^(2)))=(1)/(4ab)`
when `(x)/(2b^(3)) = (1)/(4ab) implies x = (2b^(3))/(4ab) implies x = (b^(2))/(2a)`
and `(y)/(2b(2a^(2) + b^(2))) = (1)/(4ab) implies y = (2b(2a^(2)+b^(2)))/(4ab) implies y = (2a^(2)+b^(2))/(2a)`
Hence, `{:(x = (b^(2))/(2a)),(y = (2a^(2)+b^(2))/(2a)):}}` is the required solution.