@MarkFL (1/8)^-4/3 SO This time I didn't make a mistake on the question hahaha do I make it a cube root ?

I would first use: 1/8 = 2^(-3) like so: (1/8)^(-4/3) = (2^(-3))^(-4/3) And now multiply the exponents: (1/8)^(-4/3) = (2^(-3))^(-4/3) = 2^4 = 16

1/8 = 2^(-3) < I don't understand how you get this any chance you could break that down. From a previous I'd root with the denominator [(ex. 3/2) Square root (4)= 2^3)] and use the numerator as the exponent outside the root

We have: 1/8 Now, if we recognize that 8 is the cube of 2, that is 8 = 2^3, then we can write: 1/8 = 1/(2^3) Now, when an exponent is down in the denominator, we can bring that term up into the numerator by negating the exponent 1/8 = 1/(2^3) = 2^(-3) We can think of this as: 1/(2^3) = (2^0)/(2^3) = 2^(0 - 3) = 2^(-3)

Yes, there's lots of ways to go about it...I try to show the method that I think will be the most useful in general.

Hey @MarkFL, When there are exponents inside a root, you divided by the root? example 4root (x^8y^12) would be x^2y^3 right? Or am I doing it wrong?

The short story here is that we started out with an even exponent on y and ended up with an odd exponent, and when a negative number has an odd exponent, the result is a negative value. But, when a negative number has an even exponent, the result is positive So, if y is negative, the original expression is positive, and we need to insure the end result is too.