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`10x^3y^-2` over `15x^6y^-3` This whole problem is then to the negative third
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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)`
Posted by ishpiro on June 25, 2013 at 3:02 PM (Answer #1)
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