Factorize each of the following expressions : `x^(12)-y^(12)` (ii) `x^9-y^9`

Explanation

"Solution
`x^12 - y^12`
= `(x^6)^2-(y^6)^2`
= `(x^6 +y^6) (x^6− y^6)`
=`(x^6-y^6)`
= `(x^3)^2 – (y^3)^2`
= `(x^3 +y^3) (x^3 - y^3)`
= `(x + y) (x^3 − y^3) (x^2 − xy + y^2)`
= `(x + y) (x − y) (x^2 − xy + y^2) (x^2 + xy + y^2)`
= `(x + y) (x−y) (x^6 + y^6) (x^2 −xy + y^2) (x^2 + xy + y^2)`
`:.x^12-y^12 = (x+y) (x−y) (x + y) (x^2 - xy + y^2) (x^2 + xy + y^2)`
Given(ii) `x^9-y^9`
Apply factoring rule : `x^n − y^n = (x − y) (x^n-1 + x^n−y^2+xy^n-2+xy^n-1)`
`x^9 − y^9 = (x − y) (x^8 + x^7y +x^6y^2 + x^5y^3 +x^4y^4+ +x^3y^5 + x^2y^6 +xy^7+y^8)`
`:.x^9-y^5 : (x−y) (x^8 + x^7y + x^6y^2 + x^5y^3 +x^4y^4+ + x^3y^5 + x^2y^6 + xy^7 +y^8)`