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# factorise completely to the lowest indez a^6-b^6 watch

1. factorise completely to the lowest indez a^6-b^6

can anyone help me with this problem, i think i have a solution that multiplies out ok ill varyfy with you on pm if you like, thanks
2. Difference of squares: x^2 - y^2 = (x + y)(x - y)

Replace x by a^3 and y by b^3:

a^6 - b^6 = (a^3 + b^3)(a^3 - b^3)

But

a^3 + b^3 = (a + b)(a^2 - ab + b^2),
a^3 - b^3 = (a - b)(a^2 + ab + b^2).

So

a^6 - b^6 = (a + b)(a^2 - ab + b^2)(a - b)(a^2 + ab + b^2).
3. (Original post by atkelly)
factorise completely to the lowest indez a^6-b^6

can anyone help me with this problem, i think i have a solution that multiplies out ok ill varyfy with you on pm if you like, thanks
a^6-b^6 = (a^3-b^3)(a^3+b^3) = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)

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