Basic Math Examples

\frac{64 c^{3}}{16 c^{2} + 4 c n + n^{2}} - \frac{n^{3}}{16 c^{2} + 4 c n + n^{2}}

$\frac{64 c^{3}}{16 c^{2} + 4 c n + n^{2}} - \frac{n^{3}}{16 c^{2} + 4 c n + n^{2}}$

Step 1

Combine the numerators over the common denominator.

\frac{64 c^{3} - n^{3}}{16 c^{2} + 4 c n + n^{2}}

$\frac{64 c^{3} - n^{3}}{16 c^{2} + 4 c n + n^{2}}$

Step 2

Simplify the numerator.

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Step 2.1

Rewrite

64 c^{3}

$64 c^{3}$ as

{(4 c)}^{3}

${(4 c)}^{3}$ .

\frac{{(4 c)}^{3} - n^{3}}{16 c^{2} + 4 c n + n^{2}}

$\frac{{(4 c)}^{3} - n^{3}}{16 c^{2} + 4 c n + n^{2}}$

Step 2.2

Since both terms are perfect cubes, factor using the difference of cubes formula,

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

a = 4 c

$a = 4 c$ and

b = n

$b = n$ .

\frac{(4 c - n) ({(4 c)}^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}

$\frac{(4 c - n) ({(4 c)}^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Apply the product rule to

4 c

$4 c$ .

\frac{(4 c - n) (4^{2} c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}

$\frac{(4 c - n) (4^{2} c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}$

Step 2.3.2

Raise

4

$4$ to the power of

2

$2$ .

\frac{(4 c - n) (16 c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}

$\frac{(4 c - n) (16 c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}$

\frac{(4 c - n) (16 c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}

$\frac{(4 c - n) (16 c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}$

\frac{(4 c - n) (16 c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}

$\frac{(4 c - n) (16 c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}$

Step 3

Cancel the common factor of

16 c^{2} + 4 c n + n^{2}

$16 c^{2} + 4 c n + n^{2}$ .

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Step 3.1

Cancel the common factor.

$\frac{(4 c - n) (16 c^{2} + 4 c n + n^{2})}{16 c^{2} + 4 c n + n^{2}}$

Step 3.2

Divide

$4 c - n$ by

$1$ .

$4 c - n$