1. Simplification of Fractional Expressions

How do I simplify these fractional expressions?

$\displaystyle (x + h)^-3 - x^-3 \over h$

$\displaystyle \sqrt{1+(x^3-\frac{1}{4x^3})^2}$

thanks!

2. Originally Posted by nathan02079
$\displaystyle (x + h)^-3 - x^-3 \over h$
$\displaystyle \frac{ (x + h)^{-3} - x^{-3}}{h}$

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

Multiply the numerator and denominator of the overall fraction by the LCD of $\displaystyle (x + h)^3$ and $\displaystyle x^3$:
$\displaystyle = \frac{ \frac{1}{(x + h)^3} - \frac{1}{x^3}}{h} \cdot \frac{x^3(x + h)^3}{x^3(x + h)^3}$

$\displaystyle = \frac{x^3 - (x + h)^3}{hx^3(x + h)^3}$

Now expand the $\displaystyle (x + h)^3$ in the numerator and go from there.

-Dan

3. Originally Posted by nathan02079
$\displaystyle \sqrt{1+(x^3-\frac{1}{4x^3})^2}$
$\displaystyle \sqrt{1+ \left ( x^3-\frac{1}{4x^3} \right )^2}$

$\displaystyle = \sqrt{1+ \left ( x^6 - 2 \frac{x^3}{4x^3} + \frac{1}{16x^6} \right )}$

$\displaystyle = \sqrt{x^6 + 2 \frac{x^3}{4x^3} + \frac{1}{16x^6}}$
In this last line I worked out $\displaystyle 1 - 2\frac{x^3}{4x^3} = \frac{4x^3}{4x^3} - \frac{2x^3}{4x^3} = \frac{2x^3}{4x^3}$ instead of simply making it $\displaystyle \frac{1}{2}$. This is so I can factor in the next step.

So
$\displaystyle \sqrt{1+ \left ( x^3-\frac{1}{4x^3} \right )^2} = \sqrt{x^6 + 2 \frac{x^3}{4x^3} + \frac{1}{16x^6}}$

$\displaystyle = \sqrt{\left ( x^3 + \frac{1}{4x^3} \right )^2} = \pm \left ( x^3 + \frac{1}{4x^3} \right )$

-Dan

4. Originally Posted by topsquark
$\displaystyle \frac{ (x + h)^{-3} - x^{-3}}{h}$

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

Multiply the numerator and denominator of the overall fraction by the LCD of $\displaystyle (x + h)^3$ and $\displaystyle x^3$:
$\displaystyle = \frac{ \frac{1}{(x + h)^3} - \frac{1}{x^3}}{h} \cdot \frac{x^3(x + h)^3}{x^3(x + h)^3}$

$\displaystyle = \frac{x^3 - (x + h)^3}{hx^3(x + h)^3}$

Now expand the $\displaystyle (x + h)^3$ in the numerator and go from there.

-Dan
like this?

$\displaystyle x^3-(x+h)(x+h)(x+h)\over hx^3(x+h)^3$

$\displaystyle x^3-(x^3+2hx^2+xh^2+hx^2+2h^2x+h^3)\over hx^3(x+h)^3$

$\displaystyle x^3-(x^3+3hx^2+3h^2x+h^3)\over hx^3(x+h)^3$

$\displaystyle -3hx^2-3h^2x+h^3\over hx^3(x+h)^3$

am i going in the right direction here?

5. am i going in the right direction here?
Looking a-okay to me!

6. Originally Posted by topsquark
$\displaystyle = \sqrt{\left ( x^3 + \frac{1}{4x^3} \right )^2} = \left| x^3 + \frac{1}{4x^3} \right|$

-Dan
Maybe?

7. Originally Posted by DivideBy0

$\displaystyle = \sqrt{\left ( x^3 + \frac{1}{4x^3} \right )^2} = \left| x^3 + \frac{1}{4x^3} \right |$

Maybe?
Oh, probably. I always forget when to use principle value and when not to for some reason. (Which is embarrassing since I used to teach this. )

-Dan