1. Find HCF(GCD), LCM of `(x^2-y^2),(x^3-y^3)`

Solution:
1. Factor of `(x^2-y^2)`

`=(x^2-y^2)`

`=x^2-y^2`

Rewrite `x^2=(x)^2` and `y^2=(y)^2`

`=(x)^2-(y)^2`

Here both terms are perfect squares, so factor using the difference of squares formula, `a^2-b^2=(a-b)(a+b)` where `a=x` and `b=y`

`=(x-y)(x+y)`

= ( x - y)( x + y)



 2. Factor of `(x^3-y^3)`

`=(x^3-y^3)`

`=x^3-y^3`

Rewrite `x^3=(x)^3` and `y^3=(y)^3`

`=(x)^3-(y)^3`

Here both terms are perfect cubes, so factor using the difference of cubes formula, `a^3-b^3=(a-b)(a^2+ab+b^2)` where `a=x` and `b=y`

`=(x-y)(x^2+xy+y^2)`

= ( x - y)( x² + xy + y²)


GCD = ( x - y)

LCM = ( x - y)( x + y)( x² + xy + y²)