# factorising

• Sep 13th 2009, 08:46 AM
mark
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
• Sep 13th 2009, 08:49 AM
e^(i*pi)
Quote:

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
• Sep 13th 2009, 09:06 AM
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?
• Sep 13th 2009, 09:08 AM
e^(i*pi)
Quote:

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\$
• Sep 13th 2009, 09:10 AM
mark
ah i see thanks, silly mistake there