`10x^3y^-2`  over `15x^6y^-3`  This whole problem is then to the negative third



1 Answer | Add Yours

ishpiro's profile pic

Posted on (Answer #1)

Translating this into math, we need to simplify


Numerical coefficients have a common factor of 5:

`10/15 = 2/3`

Powers of x and y can be simplified using the rules of exponents: when dividing powers with the same base, subtract exponents.

`x^3/x^6 = 1/x^3`

`y^(-2)/y^(-3) = y^1 = y`

Putting this back together, we get


Now recall then when taking negative power of a quotient, we can simply flip numerator and denominator:

`((2y)/(3x^3))^(-1/3) = ((3x^3)/(2y))^(1/3)`

Now take the power 1/3 of each factor. Keep in mind that `(x^3)^(1/3) = x` 


`((3x^3)/(2y))^(1/3) = (3^(1/3)*x)/(2^(1/3)*y^(1/3)) = (root(3)(3)*x)/root(3)(2y)`



We’ve answered 287,822 questions. We can answer yours, too.

Ask a question