factorising

**mark** · September 13th 2009, 08:46 AM

i've now got the question: factorise: $x^3 - a^3$ but i don't know what to do when its only letters involved without the help of any numbers, could someone show me how to do it this way?

thanks

**e^(i*pi)** · September 13th 2009, 08:49 AM

Originally Posted by mark

i've now got the question: factorise: $x^3 - a^3$ but i don't know what to do when its only letters involved without the help of any numbers, could someone show me how to do it this way?

thanks

This is a standard factorisation known as the difference of two cubes. Normally a and b are used but I picked p and q since a is used in the question.

$p^3 - q^3 = (p - q)(p^2 + pq + q^2)$

In your example p = x and q = a

**mark** · September 13th 2009, 09:06 AM

i tried that formula out but it didn't seem to work. say i put p = 4 and q = 2 the equation would be $4^3 - 2^3 = 64 - 8 = 56$ so then it would be $(4 - 2)(4^2 + 4(-2) + -2^2)$ which then leads to $2(16 - 8 + 4)$ then 32 - 16 + 8 = 24 ,so obviously not 56. whats gone wrong here?

**e^(i*pi)** · September 13th 2009, 09:08 AM

Originally Posted by mark

i tried that formula out but it didn't seem to work. say i put p = 4 and q = 2 the equation would be $4^3 - 2^3 = 64 - 8 = 56$ so then it would be $(4 - 2)(4^2 + 4(-2) + -2^2)$ which then leads to $2(16 - 8 + 4)$ then 32 - 16 + 8 = 24 ,so obviously not 56. whats gone wrong here?

You treated $q$ as $-q$ . The sign is already taken into account and it's treated as positive

$4^3 - 2^3 = (4-2)(4^2+4(2)+(2)^2) = 2(16+8+4) = 56$

**mark** · September 13th 2009, 09:10 AM

ah i see thanks, silly mistake there

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