Fraction with negative exponent fraction.

Question

Q: $$\left(\frac{27 a^6 b^{-3}}{c^{-2}}\right)^{-2/3}$$

A: $$\frac{b^2}{9 a^4 c^{4/3}}$$

How in the world are they getting that?

Answer 1

$$\left(\frac{27 a^6 b^{-3}}{c^{-2}}\right)^{-2/3}=\frac{(27^{-2/3})(a^{(6)(-2/3)})(b^{(-3)(-2/3)})}{c^{(-2)(-2/3)}}=\frac{b^2}{27^{2/3}a^{12/3}c^{4/3}}=\frac{b^2}{(27^{1/3})^2a^4 c^{4/3}}$$ $$=\frac{b^2}{(3)^2a^4 c^{4/3}}=\frac{b^2}{9a^4 c^{4/3}}$$

Answer 2

For one thing, $$\frac{27a^6b^{-3}}{c^{-2}}$$ is equal to $$\frac{3^3a^6c^2}{b^3}.$$

Putting that to the power of $-\frac23$ is the same as first putting it to the power of $-1$, then to the power of $2$, then to the power of $\frac{1}{3}$, so

$$\left(\frac{27a^6b^{-3}}{c^{-2}}\right)^{-\frac23} = \left(\left(\left(\frac{3^3a^6c^2}{b^3}\right)^{-1}\right)^2\right)^{\frac13}.$$

Now, simply collaps the parentheses from the inside out.

Answer 3

You just distribute the $-2/3$ power to each part, as $a^x\cdot b^x=(a\cdot b)^x$.

$$\left(\frac{27 a^6 b^{-3}}{c^{-2}}\right)^{-2/3}=\frac{27^{-2/3}\cdot (a^6)^{-2/3}\cdot (b^{-3})^{-2/3}}{(c^{-2})^{-2/3}}$$

Then just use the identity $(a^x)^y=a^{x\cdot y}$.

$$\frac{27^{-2/3}\cdot (a^6)^{-2/3}\cdot (b^{-3})^{-2/3}}{(c^{-2})^{-2/3}}=\frac{27^{-2/3}\cdot a^{-4}\cdot b^{2}}{c^{4/3}}=\frac{b^2}{9a^4c^{4/3}}$$

Answer 4

Try this way:

$$\left(\frac{27 a^6 b^{-3}}{c^{-2}}\right)^{-2/3}=(\frac{c^{-2}}{27 a^6 b^{-3}})^{2/3}=\frac{c^{-4/3}}{9a^4b^{-2}}=\dots$$

Answer 5

This fraction can be written as $$\frac{27^\frac{-2}{3}\cdot a^{(6)(\frac{-2}{3})}\cdot b^{(-3)(\frac{-2}{3})}}{c^{(-2)(\frac{-2}{3})}}$$

Or, $$\frac{3^{-2}\cdot a^{-4}\cdot b^2}{c^{\frac{4}{3}}}$$

Or, $$\frac{b^2}{9\cdot a^4\cdot c^{\frac{4}{3}}}$$

PS. I am still not perfect with TEX so, pardon me for the formatting. :)