The answer is \(\frac{x^{4/3}y^{3}}{3}\).
The fraction is: \(\frac{4xy^{-2} }{12 x^{-1/3} y^{-5} }\)
Let's rewrite it: \(\frac{4xy^{-2} }{12 x^{-1/3} y^{-5} }=\frac{4 }{12 }*\frac{x }{ x^{-1/3} }*\frac{y^{-2} }{ y^{-5} }\)
Since \(\frac{x^{a} }{x^{b}} =x^{a-b}\),
then:
\(\frac{x}{ x^{-1/3} }= x^{1-(-1/3)} = x^{3/3+1/3} = x^{4/3}\)
\(\frac{y^{-2} }{ y^{-5} }= y^{-2-(-5)} = y^{-2+5} = y^{3}\)
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\(\frac{4 }{12 }*\frac{x }{ x^{-1/3} }*\frac{y^{-2} }{ y^{-5} }= \frac{1*4}{3*4}* x^{4/3}*y^{3}= \frac{1}{3} * x^{4/3}*y^{3}= \frac{x^{4/3}y^{3}}{3}\)
