Struggling to simplify $w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ to $-w\sqrt{2w}$ I'm asked to simplify $w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ and am provided with the solution: $-w\sqrt{2w}$
I arrived at $9\sqrt{2}$ but I think I'm confused in understanding communitive rule here.
Here is my working:
$w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ = $\sqrt{w^3}\sqrt{32}$ - $\sqrt{w^3}\sqrt{50}$ # is this correct approach? I made the radical exponent a radical
Then:
$\sqrt{32}$ = $\sqrt{4}$ * $\sqrt{4}$ * $\sqrt{2}$ = $2 * 2 * \sqrt{2}$ = $4\sqrt{2}$
$\sqrt{50}$ = $\sqrt{2}$ * $\sqrt{25}$ = $5\sqrt{2}$
So:
$\sqrt{w^3}$$4\sqrt{2}$ - $\sqrt{w^3}5\sqrt{2}$ # should the expressions on either side of the minus sign be considered a single factor? i.e. could I also write as ($\sqrt{w^3}$$4\sqrt{2}$) - ($\sqrt{w^3}5\sqrt{2}$) )?
Then I'm less sure about where to go next. Since I have a positive $\sqrt{w^3}$ and a negative $\sqrt{w^3}$ I cancelled those out and was thus left with $9\sqrt{2}$.
More generally I was not sure of how to approach this and could not fin a justification for taking the path that I did.
How can I arrive at $-w\sqrt{2w}$ per the text book's solution?

You are on the right track to simplify $ w^\frac{3}{2} \sqrt{32} - w^\frac{3}{2} \sqrt{50} $ to as far as 
$$\mathrm{(1)} \qquad \sqrt{w^3} 4 \sqrt{2} - \sqrt{w^3} 5 \sqrt{2} $$
I would make expression $ (1) $ neater and rewrite as
$$\mathrm{(2)} \qquad 4 \sqrt{w^3} \sqrt{2} - 5 \sqrt{w^3} \sqrt{2} $$
These two terms are alike, and combining the two yields
$$\mathrm{(3)} \qquad -\sqrt{w^3} \sqrt{2} $$
According to one of the properties of radicals, $ \sqrt{a} \sqrt{b} = \sqrt{ab} $. Using that property and rearranging,
$$\mathrm{(4)} \qquad -\sqrt{2w^3} $$
Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes
$$\mathrm{(5)} \qquad -\sqrt{2w w^2} $$
Simplify to get the desired result
$$ -w \sqrt{2w} $$ 
