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

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### 1 Answer

Translating this into math, we need to simplify

`((10x^3y^(-2))/(15x^6y^(-3)))^(-1/3)`

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

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

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`

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