MHB Factoring when exponent is a variable

[Vac1]

I am trying to factor the following equation

$$\large(x^{\frac{1}{n}}+a)^{n-1}$$

but the fact that the exponent is n-1 is throwing me off. How could I go about factoring out this equation? Thanks.

 

[SuperSonic4]

I am trying to factor the following equation

$$\large(x^{\frac{1}{n}}+a)^{n-1}$$

but the fact that the exponent is n-1 is throwing me off. How could I go about factoring out this equation? Thanks.

Use the law of exponents to simplify the exponent: $$ \displaystyle \left(x^{\frac{1}{n}}\right)^{n-1} = \left(x^{\frac{1}{n}}\right)^n \cdot \left(x^{\frac{1}{n}}\right)^{-1} $$

The only real advantage I can see to simplifying this expression is to be able to use the binomial theorem on the exponent.

 

[Vac1]

The reason I'm trying to factor it is so that I can divide it by x.

$$\large\frac{x}{(x^{\frac{1}{n}}+a)^{n-1}}$$

I'm not concerned with the remainder at all but I'm looking for a result that should be equal to:

$$\large x^{\frac{1}{n}}-((n-1)a)$$

I'm assuming to get this, I'll need to ignore some of the terms in the polynomial when factoring which have less of an influence. Thanks

 

[Amer]

I do not know if this could help you
let x^{\frac{1}{n}} + a = u

\frac{(u-a)^n}{u^{n-1}}
but you still need the binomial theorem as what supersonic said
if you want to ignore the terms with the denominator different from 1
\frac{(u-a)^n}{u^{n-1}}\approx u - na
sub u value and factor -a

 

[caffeinemachine]

I am trying to factor the following equation

$$\large(x^{\frac{1}{n}}+a)^{n-1}$$

but the fact that the exponent is n-1 is throwing me off. How could I go about factoring out this equation? Thanks.

What do you mean by "factoring" this expression(not equation). ?? I don't quite understand what exactly you want to do with this expression.
Factoring to me is something like $x^2+5x+6=(x+2)(x+3)$

 

[SuperSonic4]

What do you mean by "factoring" this expression(not equation). ?? I don't quite understand what exactly you want to do with this expression.
Factoring to me is something like $x^2+5x+6=(x+2)(x+3)$

Factoring is essentially splitting an expression into terms each with a lower degree than the original. You could factor 8 as (2)(2)(2) or (2)(4) if you so wanted (this mainly comes up in prime factoring for LCM and GCF).
Alternatively you could factor $x^3-x = x(x^2-1) = x(x-1)(x+1)$

In this case factoring is unusual and the only way I could see it being used is to then use the binomial theorem with appropriate truncation