Pascals Triangle

[Mathysics]

how do i find the independent of x in (6x^(2)-3/x)^6

I know that the formula nCr a^(n-r) b^r

i have got an answer of n = 6 and r = 4 i just wanna make sure that im correct and i would like someone to give me a hand~ :)

[loveequation]

I don't think people (including myself) understand what you mean by finding "the independent of x".

[HallsofIvy]

Do you mean possibly the "constant term" (which is "independent of x") when you expand the polynomial? If so, then yes, r= 4 is correct. However, in future, it would be good to show how you got your answer so (just in case you are wrong!) we could indicate where you went wrong.

[Elucidus]

I believe the sought expression is "coefficient". As in: Find the coefficient of x in the expansion of the expression (6x^2-\frac{3}{x})^6.

Obviously n = 6 here. In this case one would need to find what value of r results in a^r b^{(n-r)} being a multiple of x (here a=6x^2,b=-\frac{3}{x}.

--Elucidus

[Mathysics]

I don't think people (including myself) understand what you mean by finding "the independent of x".
independent of x simply means that x^0 and yes I am trying to find the co-efficient there

heres are my working out.
independent of x in (6x^(2)-3/x)^6

1. (6^(6-r) x^(12-2r) ((-3^(r))/x^(r))

2. x^(12-2r)/x^(r) = x^(0)

3. x^(12-3r)=x^(0)

4. 12-3r=0

5. r = 4

[Elucidus]

independent of x simply means that x^0 and yes I am trying to find the co-efficient there

heres are my working out.
independent of x in (6x^(2)-3/x)^6

1. (6^(6-r) x^(12-2r) ((-3^(r))/x^(r))

2. x^(12-2r)/x^(r) = x^(0)

3. x^(12-3r)=x^(0)

4. 12-3r=0

5. r = 4

Ah, I see. This is a term I was unfamiliar with, but I understand now. Your work seems correct so far. If r = 4, what would the independent coefficient (constant term) be?

--Elucidus

[Mathysics]

thx for the help :)

exams finally over