c+66/c^2/1+4/c

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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.
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Thank you :)

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Other answers:

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 \(\dfrac{c}{1}\)
We get : \[\frac{\dfrac{c^2}{c^2} \cdot c+\dfrac{64}{c^2}}{1+\dfrac{4}{c}}\]
The numerator will then simplify to \[\frac{\dfrac{c^3 +64}{c^2}}{1+\dfrac{4}{c}}\] Are you with me so far?
yes!
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: \(\dfrac{c+4}{c}\)
We're going to take \(\dfrac{c+4}{c}\) and replace \(1+\dfrac{4}{c}\) with this simplification :) \[\frac{\dfrac{c^3+64}{c^2}}{\dfrac{c+4}{c}}\]
gotcha
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: \(\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} \iff \dfrac{a}{b} \cdot \dfrac{d}{c} \)
Therefore: \[\frac{c^3+64}{c^2} \cdot \frac{c}{c+4}\]
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?

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