Originally Posted by
Greengoblin Ok well first the minus 1 exponent in the outer brackets tells you to take the inverse of the fraction. i.e:
$\displaystyle \left(\frac{\left[3 \sqrt{27x^6y^{14}}\right]^2}{\left[\sqrt{9x^2y^6}\right]^3}\right)^{-1}=\frac{\left[\sqrt{9x^2y^6}\right]^3}{\left[3\sqrt{27x^6y^{14}}\right]^2}$
next, you can cancel the square root on the denominator, and the numerator exponent becomes 3/2, since the square root can be replaced by the exponent, 1/2, and we know as a rule of indicies, $\displaystyle (a^m)^n=a^{mn}$....
$\displaystyle \frac{\left[9x^2y^6\right]^{3/2}}{3^227x^6y^{14}}$
Next step is to use this rule again on the numerator:
$\displaystyle \frac{9^{3/2}x^{6/2}y^{18/2}}{243x^6y^{14}}=\frac{27x^3y^9}{243x^6y^{14}}=\f rac{1}{9x^3y^5}$