1. ## factorising

i've now got the question: factorise: $\displaystyle x^3 - a^3$ 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

2. Originally Posted by mark
i've now got the question: factorise: $\displaystyle x^3 - a^3$ 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.

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

In your example p = x and q = a

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

4. 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 $\displaystyle 4^3 - 2^3 = 64 - 8 = 56$ so then it would be $\displaystyle (4 - 2)(4^2 + 4(-2) + -2^2)$ which then leads to $\displaystyle 2(16 - 8 + 4)$ then 32 - 16 + 8 = 24 ,so obviously not 56. whats gone wrong here?
You treated $\displaystyle q$ as $\displaystyle -q$. The sign is already taken into account and it's treated as positive

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

5. ah i see thanks, silly mistake there