Melody

$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}$$

   $$\mbox{The (r+1)th term will be }\quad 0\le{r}\le10\quad r\in \mathBb{Z}$$      

$$10Cr\times \left(\frac{3}{x^2}\right)^r\times (-4x^3)^{10-r}\\


=10Cr \times 3^r\times(-4)^{10-r}\times \dfrac{1}{x^{2r}}\times (x^3)^{10-r}\\

=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r}\times x^{30-3r}\\

=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r+30-3r}\\

=10Cr \times 3^r\times(-4)^{10-r}\times x^{30-5r}$$ 

You want the constant term so 30-5r=0,   r=6  

so the term we want is 

$$10C6\times 3^6\times (-4)^4 \times x^0$$

etc (correct assuming I didn't make any stupid errors)

Thanks Sportslover,

Anon, 

Can you give an example of what you need to do.  Your question is a bit vague.

okay then.

do you have to pass a supervise test for that?

Are you getting us to do all your homework for you?

You have a lot of questions up here.

You are very welcome.

I see you only joined yesterday.

Welcome to Web2.0calc forum.  I hope that you learn lots and have fun here.

$$(3y-2x)^5$$

the highestr power of y is going to be 5 followed by 4,3,2,1,0

The forth largest is the power of 2 so you need the term with y² in it.  Which will be

$$(3y)^2...$$

The whole term will be

 $$5C2\times (3y)^2 \times (-2x)^3 = 10\times9y^2\times -8x^3 = -720x^3y^2$$

Now if you don't understand, tell me what is troubling you and I will attempt to explain better!

That is what I am saying.  you should ask people to explain.  And give thumbs up when you are totally satisfied that you have been taught or at lest told what you need to know.

would you like me to try and explain or are you happy with just the answer?

weren't you supposed to be able to do it without a calculator?

Alan, I strong suspect that this was supposed to be done without a calculator!

Although, since the Sportslover gave you a thumbs up, maybe I am wrong.

$$\dfrac{3}{x^2}-4x^3$$   

There is not expansion here and no constant.  You had best check the question.