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Dont know how to solve (n^3-1)/(n^2-1)
- Thread starter Zulgok
- Start date
D
Deleted member 4993
Guest
(n^3-1)/(n^2-1)
Im not sure how to solve it, I only know what could be done to the lower part of the fraction, but no idea for the upper part. Is it something like (n-1)(n-1)(n+1) and the lower part (n-1)(n+1) ? Thank you.
Use:
a3 - b3 = (a - b)(a2 + ab + b2)
D
Deleted member 4993
Guest
Ok, thank you. You also dont need that "2" anymore in (a-b)(a^2+(2?)ab+b^2) ?
You need that two (2) for entirely different equation:
(a + b)2 = a2 + 2ab + b2
D
Deleted member 4993
Guest
Also, another question. If discriminant is less than 0, what do I do with quadratic equation ? Instead of ax^2+bx+c=(x-x1)(x-x2) should I write =(x-0)(x-0) or something ?
Unless you are allowed to use complex numbers, leave it alone (cannot be factorized).