limits: need to factor denominator in (m - 1)/(m^6 - 1)

[Tascja]

when solving a limit by changing the variable via substitution, it ends up being:

m-1 / m^6-1

how do i factor m^6-1??

 

[o_O]

Some formulae that might help you:
\(\displaystyle a^{2} - b^{2} = (a - b)(a+b)\) (Imagine a² = m⁶ and b² = 1)

\(\displaystyle a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\)

\(\displaystyle a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\)

Show us your work and we can locate any errors for you.

 

[Tascja]

so for my question it would be:

m-1/ m^6-1
= m-1/ (m^3-1)(m^3+1)
=m-1/ (m-1)(m^2+m+1)(m^3+1)

which then (m-1) cancels with the top to be left with:
= 1/ (m^2+m+1)(m^3+1)

 

[o_O]

Yep. Don't know what you want to do with that though.

 

[morson]

Which can be further factorised to:

\(\displaystyle \L\ \frac{1}{(m + 1)(m^2 + m + 1)(m^2 - m + 1)}\\)

So long as m \(\displaystyle \L\ \not=\ 1\)