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

1.	Factorise : `(a) x^(6)+y^(6) " " (b) x^(6)-y^(6)`
Answer» (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))`

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