1. ## simplify

This will most likely be hard to interpret, I apologize.

$\displaystyle (3 \sqrt{(27x^6y^14)}^2/ \sqrt{(9x^2y^6)}^3)^-1$

Note: the fraction is in brackets, therefore it is all to the power of -1.

2. Do you mean $\displaystyle \sqrt{(27x^6y^{14})^2}$ and $\displaystyle \sqrt{(9x^2y^6)^3}$ , or $\displaystyle \left[\sqrt{(27x^6y^{14})}\right]^2$ and $\displaystyle \left[\sqrt{(9x^2y^6)}\right]^3$?

3. Originally Posted by Greengoblin
Do you mean $\displaystyle \sqrt{(27x^6y^{14})^2}$ and $\displaystyle \sqrt{(9x^2y^6)^3}$ , or $\displaystyle \left[\sqrt{(27x^6y^{14})}\right]^2$ and $\displaystyle \left[\sqrt{(9x^2y^6)}\right]^3$?
$\displaystyle \left[3\sqrt{(27x^6y^{14})}\right]^2$ and $\displaystyle \left[\sqrt{(9x^2y^6)}\right]^3$

Note the 3 before the sqrt in the numerator and all is to the power of -1, such that the ^2 in the numerator and the ^3 in the denominator are inside the brackets.

4. 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:

$\displaystyle \frac{9^{3/2}x^{6/2}y^{18/2}}{243x^6y^{14}}=\frac{27x^3y^9}{243x^6y^{14}}=\f rac{1}{9x^3y^5}$

5. 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:

$\displaystyle \frac{9^{3/2}x^{6/2}y^{18/2}}{243x^6y^{14}}=\frac{27x^3y^9}{243x^6y^{14}}=\f rac{1}{9x^3y^5}$

I really appreciate the explanations, it helps very much.