# Binomial Expansion

1. Oct 16, 2005

### zbobet2012

I understand how Binomial expansion works, but I don't understand how to solve this problem.
Give the term of (2/x^2-x)^6 that has no x.

2. Oct 16, 2005

### robert Ihnot

I take it you mean: $$(\frac{2}{X^2}-X)^6$$. In this case we simply want to solve 2A=6-A for the term where (-X)^(6-A) and X^2 is raised to the term A. Obviously, A=2, giving: $$\frac{6!}{4!2!}2^2$$

Last edited: Oct 17, 2005
3. Oct 16, 2005

### Tide

If a = 2/x^2 and b = -x then the expansion will contain various products of powers of a and b. Some of those products will be such that the x's cancel. Can you see which ones? Can you calculate their coefficients using the Binomial theorem?

4. Oct 17, 2005

### zbobet2012

Where did this: 2A=6-A

Come from?

5. Oct 17, 2005

### Tide

Robert means that each term in the series will be of degree 6, i.e. the combined powers of a and b (from my earlier post) add up to 6. For one or more of those terms the power of x will be zero.

6. Oct 17, 2005

### zbobet2012

I know its asking alot, but can you show a step by step on how to solve it? I was out of class for a few days and never got taught how... Thanks alot.

7. Oct 17, 2005

### Tide

You said you understood how the binomial expansion works so you can easily do it yourself.

Expand $(a + b)^6$ using the binomial expansion. As a shortcut, you can use Pascal's Triangle to find the binomial coefficients. When you're done with that, replace a with $2/x^6$ and b with $-x$. Your answer should then leap off the page!

Good luck.

8. Oct 17, 2005

### zbobet2012

Thanks alot tide, that helps alot. I also think I found a generalized method for finding the $$x^n$$ term.
If we have $$(\frac{C}{X^k}-X^m)^z$$ than $$x^n$$ can be found where $$(-x)^{z-a}$$ where $$ka=z-ma+n$$

Last edited: Oct 17, 2005