How to simplify this fraction with different powers?

Question

I happen to be stuck trying to simplify this: $$\left[\frac{(3x+2)(x+1)^\frac{3}{2}-(\frac{3}{2}x^2+2x)(\frac{3}{2})(x+1)^\frac{1}{2}}{(x+1)^3}\right]$$

here's the simplified solution that I'm trying to figure out how it was reached

Answer 1

We have

$$\frac{(3x+2)(x+1)^\frac{3}{2}-(\frac{3}{2}x^2+2x)(\frac{3}{2})(x+1)^\frac{1}{2}}{(x+1)^3}$$

divide numerator and denominator by $(x+1)^{\frac12}$ giving

$$\frac{(3x+2)(x+1)-(\frac{3}{2}x^2+2x)(\frac{3}{2})}{(x+1)^\frac52}$$

multiply the $(3x+2)(x+1)$ to get $3x^2 + 5x + 2$ and the $(\frac32x^2 + 2x)(\frac32)$ to get $\frac94x^2 + 3x$ giving:

$$\frac{3x^2 + 5x + 2-\frac94x^2 - 3x}{(x+1)^\frac52}$$

then simplify the numerator:

$$3x^2 + 5x + 2-\frac94x^2 - 3x = \frac{12}{4}x^2 - \frac94x^2 + 5x - 3x + 2 = \frac34x^2 + 2x + 2$$

to get

$$\frac{\frac34x^2 + 2x + 2}{(x+1)^\frac52}$$

Answer 2

$$\frac{(3x+2)(x+1)^\frac{3}{2}-(\frac{3}{2}x^2+2x)(\frac{3}{2})(x+1)^\frac{1}{2}}{(x+1)^3}=$$

take the common factor $(x+1)^\frac{1}{2}$

$$=\frac{[(3x+2)(x+1)-(\frac{3}{2}x^2+2x)(\frac{3}{2})]\color{red}{(x+1)^\frac{1}{2}}}{(x+1)^3}=$$ $$=\frac{(3x+2)(x+1)-(\frac{3}{2}x^2+2x)(\frac{3}{2})}{(x+1)^3\color{red}{(x+1)^{-\frac{1}{2}}}}=\frac{(3/4)x^2+2x+2}{(x+1)^{\frac{5}{2}}}$$