Find Coefficient of x^3 in (2x^2-3/x)^3

[Stacyg]

Find the coefficient of x^3 in the expansion of (2x^2-3/x)^3


I know how to do simple coefficients using pascalles triangle but I really don't know how to do this.
Any help would be much appreciated.

 

[Tedjn]

Write it as

(2x^2 - 3x^{-1})^3

From Pascal's triangle, you know how to expand

(a+b)^n

What can you replace with a and what can you replace with b?

 

[rock.freak667]

(2x^2-\frac{3}{x})^3

(\frac{1}{x}(2x^3-3))^3

How about now?

 

[Tedjn]

Oh, that's a nice way of doing it :)

 

[rock.freak667]

Oh, that's a nice way of doing it :)

Usually (well for me), a binomial expansion is usually done with a variable and a constant.

as for (a+b)^n is valid for |\frac{b}{a}|<1 But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.

 

[Dick]

Usually (well for me), a binomial expansion is usually done with a variable and a constant.

as for (a+b)^n is valid for |\frac{b}{a}|<1 But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.

Why is it only valid in some range? I also don't see why you need to factor the original. (a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3. Just put a=2x^2 and b=(-3/x), figure out which term is the x^3 term and evaluate it.

 

[rock.freak667]

That's what I was taught.."validity of a binomial"

 

[Dick]

That's what I was taught.."validity of a binomial"

Got a reference? If you are thinking of the convergence of the infinite series for negative exponents, that is something to think about. But this is a positive exponent, the series is finite. There are no convergence issues.

 

[BrendanH]

Besides, we're dealing with polynomials in the case of (a+b)^n

 

[Dick]

Besides, we're dealing with polynomials in the case of (a+b)^n

Yeah, that's what I mean by finite series. You could also just forget about pascal's triangle and multiply it out. The power is only 3.