(C1) Binomial Expansion

Exam board: OCR MEI
Unit: C1

Question: Expand $(x-\frac{3}{x})^5$
I'm not really sure what I'm meant to be doing here... I've been told to write out $(a+b)^5$ and substitute in from the original equation.. But I'm not really sure where to go from there... I've got this;

$x^5+5x^4(-\frac{3}{x})+10x^3(-\frac{3}{x})^2+10x^2(-\frac{3}{x})^3+5x(-\frac{3}{x})^4+(-\frac{3}{x})^5$

Hopefully that makes sense.

I'm open to finding other ways to work it out, but I'd prefer it if you could help by using the method I've been given. Thanks.

The third term for example is $10x^3(-\frac{3}{x})^2=10x^3\frac{9}{x^2}=90x$

I can't see how you got from the second to the third part...

$\frac{x^3}{x^2}=x$ and $10 \times 9=90$

Ahh yes I literally got it just then. So does that mean the fourth term is $\frac{270}{x}$?

yes thats correct , but remember there is a negative sign in front of the 270/x so the answer would be = -270/x

I'm having trouble with the fifth term. I've done everything else.

$x^5 - 15x^3 + 90x - \frac{270}{x} + 5x(-\frac{3}{x})^4 -\frac{243}{x^5}$

I've got as far as $5x\frac{81}{x^4}$, and if I multiply the 5 and 81 I obviously get 405, and I was to do the same with the x's then altogether I'd get $405x^{-3}$ but that doesn't seem right... and as far as I know I can't do anything with that..

why doesnt it seem right ? 405x^-3 is also 405/x^3 , seems fine to me.

Oh yeah, I forgot that 405x^-3 would also be the same as 405/x^3.

So I put that in, and everything is correct, right?

I never did it that way, are you using nCr but just writing it the long way?

I never did it that way, are you using nCr but just writing it the long way?

I thought nCr was to find a specific term?

nCr just finds the 1, 5, 10, 10, 5, 1 so you used it here

I used Pascal's Triangle to do that... Are they the same thing?