# Thread: Complex Fraction with negative powers above -1

1. ## Complex Fraction with negative powers above -1

Hello,

If for example to simplify within a complex fraction it has (x+h)-3 is that 1/(x+h)3 or 3/(x+h)?

as (x+h)-1 is 1/(x+h)

Thanks (im pretty sure its the cubed, but just want to make sure).

The example is in my book, but its an even numbered question so the answer isn't in the back.. and there are no examples with negative powers over 1.

The actual question is:

Simplify:
$\frac{(x+h)^-3 -x^-3}{h}$

2. Originally Posted by captain_jonno
Hello,

If for example to simplify within a complex fraction it has (x+h)-3 is that 1/(x+h)3 or 3/(x+h)?

as (x+h)-1 is 1/(x+h)

Thanks (im pretty sure its the cubed, but just want to make sure).

The example is in my book, but its an even numbered question so the answer isnt in the back.. and there are no examples with negative powers about 1.

The actual question is:

$(x+h)^-3 - x^-3$ all divided by h
$(x+h)^{-3} = \frac{1}{(x+h)^3} \: , \: x \neq h$

There's not a great amount you can do with that, you could use the binomial expansion on (x+h)^3 on the bottom

$\frac{1}{h}((x+h)^{-3} - x^{-3})$

Note that this does not equal $\frac{1}{h} \times \frac{1}{(x+h)^3 - x^3}$

3. worked it out.

4. To clarify the question:

Simplify:
$\frac{(x+h)^-3 -x^-3}{h}$

5. Is your question regarding how to work with negative exponents, or how to simplify the complex fraction? Or both?

You have:

$\frac{(x\,+\,h)^{-3} \,-\,x^{-3}}{h}$

This means:

$\frac{\frac{1}{(x\, +\, h)^3}\, -\, \frac{1}{x^3}}{h}$

Multiplying top and bottom by x^3(x + h)^3, we get:

$\frac{x^3\, -\, (x\, +\, h)^3}{h x^3 (x\, +\, h)^3}$

$\frac{x^3\, -\, x^3\, -\, 3hx^2\, -\, 3h^2 x\, -\, h^3}{h x^3 (x\, +\, h)^3}$

$\frac{-h(3x^2\, +\, 3hx\, +\, h^2)}{h x^3 (x\, +\, h)^3}$

Cancel the one common factor.