Factoring when exponent is a variable

Vac

New member
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

Well-known member
MHB Math Helper
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.

Vac

New member
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

Active member
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

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caffeinemachine

Well-known member
MHB Math Scholar
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

Well-known member
MHB Math Helper
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