How to factorise this question?

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Answer 1 · 2018-02-23T00:08:02

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

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#

Answer 2 · 2018-02-23T03:24:27

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

#(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)#