Question

Factorise the following a ^3-a^2b^2-ab+b^3

Answer 1

a3+b3+(3a2b+3ab2)a3+b3+(3a2b+3ab2)

Since both terms are perfect cubes, factor using the sum of cubes formula, a3+b3=(a+b)(a2−ab+b2)a3+b3=(a+b)(a2-ab+b2) where a=aa=a and b=bb=b.

(a+b)(a2−ab+b2)+(3a2b+3ab2)(a+b)(a2-ab+b2)+(3a2b+3ab2)

Factor 3ab3ab out of 3a2b+3ab23a2b+3ab2.

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(a+b)(a2−ab+b2)+3ab(a+b)(a+b)(a2-ab+b2)+3ab(a+b)

Factor a+ba+b out of (a+b)(a2−ab+b2)+3ab(a+b)(a+b)(a2-ab+b2)+3ab(a+b).

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(a+b)(a2−ab+b2+3ab)(a+b)(a2-ab+b2+3ab)

Add −ab-ab and 3ab3ab.

(a+b)(a2+b2+2ab)(a+b)(a2+b2+2ab)

Factor using the perfect square rule.

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(a+b)(a+b)2(a+b)(a+b)2

a3+3a2b+3ab2+b3a3+3a2b+3ab2+b3

Answer 2

Answer:

(a^2-b)(a-b^2) is the answer