Finding the term with no x in (2/x^2-x)^6

[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.

 

[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: $$\frac{6!}{4!2!}2^2$$

 

[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?

 

[zbobet2012]

Where did this: 2A=6-A

Come from?

 

[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.

 

[zbobet2012]

I know its asking a lot, 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 a lot.

 

[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.

 

[zbobet2012]

Thanks a lot tide, that helps a lot. 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 $$ka=z-ma+n$$