Simplification of Fractional Expressions

**nathan02079** · December 9th 2007, 05:42 PM

How do I simplify these fractional expressions?

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

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

thanks!

**topsquark** · December 9th 2007, 06:16 PM

Originally Posted by nathan02079

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

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

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

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

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

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

-Dan

**topsquark** · December 9th 2007, 06:21 PM

Originally Posted by nathan02079

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

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

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

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

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

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

-Dan

**nathan02079** · December 9th 2007, 06:29 PM

Originally Posted by topsquark

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

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

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

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

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

-Dan

like this?

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

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

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

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

am i going in the right direction here?

**colby2152** · December 9th 2007, 06:36 PM

am i going in the right direction here?

Looking a-okay to me!

**DivideBy0** · December 9th 2007, 06:36 PM

Originally Posted by topsquark

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

-Dan

Maybe?

**topsquark** · December 10th 2007, 06:28 AM

Originally Posted by DivideBy0

$= \sqrt{\left ( x^3 + \frac{1}{4x^3} \right )^2} = \left| x^3 + \frac{1}{4x^3} \right |<br />$

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

Thread: Simplification of Fractional Expressions

LinkBack

Thread Tools

Display

Simplification of Fractional Expressions

Similar Math Help Forum Discussions

fractional exponent simplification

Rational Expressions and Radical Expressions

Fractional Equations?

Fractional Calculus

Fractional equation

Search Tags