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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

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1  
We do not know that when $m,n$ are not integers, unless $a$ is positive. –  Thomas Andrews Jan 23 '13 at 15:29
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I'm surprised no one's mentioned $-3^3=-(3^3)$, not $(-3)^3$. –  Mike Jan 23 '13 at 16:06
    
en.wikipedia.org/wiki/… –  lab bhattacharjee Jan 23 '13 at 16:43

4 Answers 4

up vote 14 down vote accepted

$\exists m,n\in\Bbb Q$ such that $(a^n)^m \not= a^{nm} $ if $a<0$.

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In fact the symbol $\neq$ is not really the right one. It implies 'is not the same as', but one needs a symbol that implies something like 'is not necessarily the same as' but I guess that one doesn't exist. –  barto Jan 23 '13 at 17:04
    
I meant that the statement is not true for all values of m and n. –  Ishan Banerjee Jan 23 '13 at 17:06

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: $$-3^3 = -(3^3) = -(3\cdot 3\cdot 3) = -27$$

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You can not do that. I mean $(-3)^1\neq\sqrt{(3)^2}$, since $-3<0$.

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nice example...+1 –  amWhy Jan 23 '13 at 15:36
    
@amWhy: Thanks so much! –  Babak S. Jan 23 '13 at 15:40

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.

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