What is wrong with this problem We know that:
$(a^n)^m=a^{nm}$
From this we have:
$-3^3=[(-3)^2]^\frac{3}{2}=(3^2)^\frac{3}{2}=27$
Find what's wrong
 A: The "law of exponents" that you cite:
$$\large (a^n)^m=a^{nm}$$ 
applies PROVIDED $\bf{a \gt 0}$. 
Here, though, we have $\,\bf{a = -3 \lt 0}$, and hence:
$$-3^3=[(-3)^2]^{\large\frac{3}{2}}=(3^2)^{\large\frac{3}{2}}=27\quad \Longleftarrow \;\text{ False}.$$
$$-3^3 = -(3^3) = -[(3^2)^{\large \frac{3}{2}}] = -(9^{\large \frac{3}{2}}) = -27\quad \Longleftarrow \; \text{ True}$$
But why the trouble? It is a very straightforward computation: 
$$-3^3 = -(3^3) = -(3\cdot 3\cdot 3) = -27$$
A: You can not do that. I mean $(-3)^1\neq\sqrt{(3)^2}$, since $-3<0$.
A: when you are increasing the power of a by 3/2, you are actually cubing a and then taking square root, and you know that when to take the square root of a number, you have to consider both positive and negative roots. In this case you are neglecting the negative root which was your answer.
Remember one thing:: If you want to solve an equation and you are increasing the power of unknown(x) then you might get some extra values of x as solutions which may not be solution of your original equation.
A: $\exists m,n\in\Bbb Q$ such that $(a^n)^m \not= a^{nm}  $ if $a<0$.
