Simplify $\frac{0.32\times 0.72\times 0.56}{(0.32{)}^3+(0.24{)}^3-(0.56{)}^3}$:

Let $a=0.32$, $b=0.24$, and $c=-0.56$.Let's check the sum $a+b+c$:
$$a+b+c=0.32+0.24+(-0.56)=0.56-0.56=0exthours$$
We know the algebraic identity:
$$\text{If }a+b+c=0,\text{ then }{a}^3+b^3+c^3=3abc$$
So, the denominator is:
$$(0.32{)}^3+(0.24{)}^3+(-0.56{)}^3=3\times (0.32)\times (0.24)\times (-0.56)$$
$$(0.32{)}^3+(0.24{)}^3-(0.56{)}^3=-3\times 0.32\times 0.24\times 0.56$$
Now rewrite the given expression:
$$\text{Expression}=\frac{0.32\times 0.72\times 0.56}{-3\times 0.32\times 0.24\times 0.56}$$
Cancel out common terms $0.32$ and $0.56$:
$$\text{Expression}=\frac{0.72}{-3\times 0.24}=\frac{0.72}{-0.72}=-1$$