## anonymous one year ago c+66/c^2/1+4/c

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1. anonymous

Do you mind drawing out your question? Im not understanding which ones are fractions and which arent. Your lack of parenthesis makes it hard to differentiate that.

2. anonymous

|dw:1444169246249:dw|

3. anonymous

Thank you :)

4. anonymous

First we need to find a common denominator in both the numerator and denominator. For the numerator, the LCD between $$1$$ and $$c^2$$ is $$c^2$$. Let's multiply $$c^2$$ to the top and bottom of $$\dfrac{c}{1}$$

5. anonymous

We get : $\frac{\dfrac{c^2}{c^2} \cdot c+\dfrac{64}{c^2}}{1+\dfrac{4}{c}}$

6. anonymous

The numerator will then simplify to $\frac{\dfrac{c^3 +64}{c^2}}{1+\dfrac{4}{c}}$ Are you with me so far?

7. anonymous

yes!

8. anonymous

Great. Now let's find the LCD in the denominator. The LCD in the denominator between $$1$$ and $$c$$ will be $$c$$, therefore we multiply $$c$$ to both the numerator and denominator of 1. $\frac{c}{c} \cdot 1 + \frac{4}{c}$ This becomes: $$\dfrac{c+4}{c}$$

9. anonymous

We're going to take $$\dfrac{c+4}{c}$$ and replace $$1+\dfrac{4}{c}$$ with this simplification :) $\frac{\dfrac{c^3+64}{c^2}}{\dfrac{c+4}{c}}$

10. anonymous

gotcha

11. anonymous

When we're dividing 2 fractions, we're going to multiply by the reciprocal of the second fraction to the first fraction. It follows this simple rule: $$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} \iff \dfrac{a}{b} \cdot \dfrac{d}{c}$$

12. anonymous

Therefore: $\frac{c^3+64}{c^2} \cdot \frac{c}{c+4}$

13. anonymous

We're going to cross cancel our like terms but if we take a look, $$c^3+64$$ can be expanded since it is of the form $$a^3+b^3$$. Do you follow?