How to factorise this question?

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2 Answers
Feb 23, 2018

(a+b)(a^2-ab+b^2)+(a-b)(a^2+ab+b^2)

Explanation:

The sum and difference between perfect cubes has a formula

Sum:
(a+b)^3
(a^3+b^3)
(a+b)(a^2-ab+b^2)

Difference:
(a-b)^3
(a^3-b^3)
(a-b)(a^2+ab+b^2)

So add them together, you get
(a+b)(a^2-ab+b^2)+(a-b)(a^2+ab+b^2)

Another way:
(a+b)^3+(a-b)^3
(a^3+b^3)+(a^3-b^3)
a^3+b^3+a^3-b^3
2a^3

Feb 23, 2018

=> 2a (a^2 + 3b^2)

Explanation:

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

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=> (a^3 + b^3 + 3ab(a+b)) +( a^3- b^3 -3ab(a-b))

=> a^3 + cancel(color(red)(b^3) )+cancel( 3a^2b ) + 3ab^2 + a^3 - cancel(color(red)(b^3 ))- cancel (3a^2b )+ 3 ab^2

2a^3 + 6ab^2 = 2a (a^2 + 3b^2)