Factorise : `(a) x^(6)+y^(6) " " (b) x^(6)-y^(6)`

Explanation

(a) We have
`x^(6)+y^(6)=(x^(2))^(3)+(y^(2))^(3)`
`=(x^(2)+y^(2))(x^(4)-x^(2)y^(2)+y^(4))`
`[because a^(3)+b^(3)=(A+b)(a^(2)-ab+b^(2))]`
Most of students write factors till here, which is not proper. We shall factorise further,
as
`x^(6)+y^(6)=(x^(2)+y^(2))[underset("making perfect square")(ubrace((x^(2))^(2)+2x^(2).y^(2)+(y^(2))^(2)))-2x^(2)y^(2)-x^(2)y^(2)]`
`implies x^(6)+y^(6)=(x^(2)+y^(2))[(x^(2)+y^(2))^(2)-3x^(2)y^(2)]=(x^(2)+y^(2))[(x^(2)+y^(2))^(2)-(sqrt(3)xy)^(2)]`
`=(x^(2)+y^(2))[(x^(2)+y^(2)+sqrt(3)xy)(x^(2)+y^(2)-sqrt(3)xy)]`
`implies underset("polynomial")(ubrace(x^(6)+y^(6)))=underset("polynomial")(ubrace((x^(2)+y^(2))))underset("polynomial")(ubrace((x^(2)+sqrt(3)xy+y^(2))))underset("polynomial")(ubrace((x^(2)-sqrt(3)xy+y^(2))))`
`(b) x^(6)-y^(6)=(x^(2))^(3)-(y^(2))3)=(x^(2)-y^(2))(x^(4)+x^(2)y^(2)+y^(4))`
`=(x+y)(x-y)[underset("making perfect square")ubrace((x^(2))^(2)+2(x^(2))(y^(2))+(y^(2))^(2))-2x^(2)y^(2)+x^(2)y^(2)]`
`=(x+y)(x-y)[(x^(2)+y^(2))^(2)-(xy)^(2)]`
`=(x+y)(x-y)[(x^(2)+y^(2)+xy)(x^(2)+y^(2)-xy)]`
`=(x+y)(x-y)(x^(2)+xy+y^(2))(x^(2)-xy+y^(2))`
Alternative Method :
`x^(6)-y^(6)=(x^(3))^(2)-(y^(3))^(2)=(x^(3)+y^(3))(x^(3)-y^(3))`
`=(x+y)(x^(2)-xy)+y^(2))(x-y)(x^(2)+xy+y^(2))`
`=(x+y)(x-y)(x^(2)+xy+y^(2))(x^(2)-xy+y^(2))`