How do you express (3x^2)^-3 with positive exponents?

Sep 11, 2016

${\left(3 {x}^{2}\right)}^{- 3} = \frac{1}{27 {x}^{6}}$

Explanation:

We can use the identities ${a}^{- m} = \frac{1}{a} ^ m$, ${\left({a}^{m}\right)}^{n} = {a}^{m n}$ and ${\left(a b\right)}^{m} = {a}^{m} {b}^{m}$

Hence, ${\left(3 {x}^{2}\right)}^{- 3}$

= $\frac{1}{3 {x}^{2}} ^ 3$

= $\frac{1}{{3}^{3} \cdot {\left({x}^{2}\right)}^{3}}$

= $\frac{1}{27 \cdot {x}^{2 \times 3}}$

= 1/(27x^6

Sep 18, 2016

$\frac{1}{27 {x}^{6}}$

Explanation:

Method Example:

Suppose we had ${t}^{- 3}$. This is another way of writing $\frac{1}{{t}^{3}}$

Suppose we had $\frac{1}{{t}^{-} 3}$ this is another way of writing ${t}^{3}$

So the negative power (index) has the effect of moving the value to the other side of the horizontal line.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:$\text{ } {\left(3 {x}^{2}\right)}^{-} 3$

The index $- 3$ is acting on everything inside the brackets. So this is the same as:

$\frac{1}{3 {x}^{2}} ^ 3 \text{ "->" "1/(3^3xx x^2 xx x^2 xx x^2)" "->" } \frac{1}{{3}^{3} {x}^{2 \times 3}}$

$= \frac{1}{27 {x}^{6}}$