Question

(c^3/c^2-8c+16) - (c^2/c-4)

Answer 1

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

Answer 2

To simplify the given expression, let's start by factoring the denominators:

1. Factor the denominator of the first fraction, c^2 - 8c + 16: c^2 - 8c + 16 = (c - 4)(c - 4) = (c - 4)^2

2. Factor the denominator of the second fraction, c - 4 (It's already factored).

Now, the expression becomes:

(c^3 / (c - 4)^2) - (c^2 / (c - 4))

Next, we need to find a common denominator for both fractions, which is (c - 4)^2, and rewrite the fractions with this common denominator:

(c^3 / (c - 4)^2) - (c^2 / (c - 4)) = (c^3 / (c - 4)^2) - ((c^2 * (c - 4)) / (c - 4)^2)

Now, the fractions have the same denominator, so we can combine them:

(c^3 - c^2 * (c - 4)) / (c - 4)^2

Next, let's simplify further by expanding the terms:

c^3 - c^2 * (c - 4) = c^3 - c^3 + 4c^2 = 4c^2

The expression now becomes:

(4c^2) / (c - 4)^2

Since there are no common factors in the numerator and the denominator, the simplification is complete, and the final expression is:

(4c^2) / (c - 4)^2