Factorize a^3-b^3 (O Level Math Tuition Question)

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$

So,
$\begin{array}{rcl} a^3-b^3&=&(a-b)^3+3a^2b-3ab^2\\ &=&(a-b)(a-b)^2+3a^2b-3ab^2\\ &=&(a-b)(a^2-2ab+b^2)+(a-b)(3ab)\\ &=&(a-b)(a^2+ab+b^2) \end{array}$

Author: mathtuition88

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5 thoughts on “Factorize a^3-b^3 (O Level Math Tuition Question)”

Paperpc@yahoo.com.sg says:

May 1, 2013 at 1:27 am

How do we get (a-b)^3=a^3-3a^2b+3ab^2-b^3 if it is not (a-b)(a-b)(a-b)

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mathtuition88 says:

May 1, 2013 at 9:47 am

Hi, yes, you are right, $(a-b)^3$ is $(a-b)(a-b)(a-b)$ .
Continue to expand, you will get $(a^2-2ab+b^2)(a-b)=a^3-2a^2b+ab^2-a^2b+2ab^2-b^3=a^3-3a^2b+3ab^2-b^3$ .
Hope it helps!

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tomcircle says:

May 12, 2013 at 1:12 am

There is one challenging USA Math Olympiad 1997 (USA) question using this technique, see
my blog at:
http://tomcircle.wordpress.com/2013/05/07/imo-technique/

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Tara@ Math Tuition says:

September 25, 2013 at 8:34 pm

Your blogs seems to me a great place for math lovers. I love to solve math problems…so this is the right place for me…Thanks for sharing o level maths here.

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mathtuition88 says:

December 22, 2014 at 7:27 pm

Reblogged this on Singapore Maths Tuition.

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