Simplify \( \frac{c+\frac{8}{c^{2}}}{1+\frac{2}{c}} \)

To simplify the expression \( \frac{c+\frac{8}{c^{2}}}{1+\frac{2}{c}} \), we start by finding a common denominator for both the numerator and the denominator. **Numerator:** The numerator is \( c + \frac{8}{c^{2}} \). The common denominator here is \( c^{2} \): \[ c = \frac{c^{3}}{c^{2}} \quad \text{(multiply the numerator by } c^{2} \text{)} \] So, \[ c + \frac{8}{c^{2}} = \frac{c^{3}}{c^{2}} + \frac{8}{c^{2}} = \frac{c^{3} + 8}{c^{2}}. \] **Denominator:** Now, for the denominator \( 1 + \frac{2}{c} \), the common denominator is \( c \): \[ 1 = \frac{c}{c} \quad \text{(multiply the numerator by } c \text{)} \] So, \[ 1 + \frac{2}{c} = \frac{c}{c} + \frac{2}{c} = \frac{c + 2}{c}. \] **Now we can rewrite the original expression:** \[ \frac{c+\frac{8}{c^{2}}}{1+\frac{2}{c}} = \frac{\frac{c^{3} + 8}{c^{2}}}{\frac{c + 2}{c}}. \] **To divide by a fraction, multiply by its reciprocal:** \[ = \frac{c^{3} + 8}{c^{2}} \times \frac{c}{c + 2} = \frac{c(c^{3} + 8)}{c^{2}(c + 2)}. \] **Simplifying further:** \[ = \frac{c^{4} + 8c}{c^{2}(c + 2)}. \] At this point, we can factor \(c^{4} + 8c\): \[ = \frac{c(c^{3} + 8)}{c^{2}(c + 2)}. \] Recognizing that \( c^{3} + 8 \) can be factored using the sum of cubes: \[ c^{3} + 8 = (c + 2)(c^{2} - 2c + 4). \] **So the full expression becomes:** \[ = \frac{c(c + 2)(c^{2} - 2c + 4)}{c^{2}(c + 2)}. \] Now, we can cancel \( (c + 2) \) (assuming \( c \neq -2 \)): \[ = \frac{c(c^{2} - 2c + 4)}{c^{2}}. \] Finally, this simplifies to: \[ = \frac{c^{2} - 2c + 4}{c}. \] Putting it all together, the simplified form of \( \frac{c+\frac{8}{c^{2}}}{1+\frac{2}{c}} \) is: \[ \frac{c^{2} - 2c + 4}{c}. \]