Can you help me with this problem please I don't even know where to begin with what's in parentheses are fractions


-28x^3y^(-1/4)
4x^(-1/5)y =

-28x^3y^(-1/4)
4x^(-1/5)y =


Remember that \Large x^{-a} = \frac {1}{x^a}

Now I hope I understood your problem correctly. It was not trivial to figure out. Next time, try using parentheses and "/" for division and ^ for exponentiation.

[-28x^3 y^(-1/4)] / [4x^(-1/5) y]

or, better, learn how to use LaTeX ;)

Post on using LaTeX (https://www.askmehelpdesk.com/math-sciences/how-technical-scientific-documentation-formulas-50415.html)

\frac {-28x^3y^{(-1/4)}}{4x^{(-1/5)}y} =

\Large \frac {-28x^3 \frac {1}{y^{(1/4)}}}{4 \frac {1}{x^{(1/5)}} y } = \frac {-28x^3 x^{(1/5)}}{4y^{(1/4)} y} = \frac {-28 x^{(16/5)}}{ 4 y^{(5/4)}} = \frac {-7 x^{\frac {16}{5}}}{ y^{\frac 54}}

\Large \frac{-28x^3y^{-\frac1 4}}{4x^{-\frac1 5}y}\;=\\
\frac{-28x^3x^{\frac1 5}}{4y^{\frac1 4}y}\;=\\
\frac{-7x^{\frac{16} 5}}{y^{\frac5 4}}

Thanks. Did you drop 4 in the denominator? I can't see where it went.

LaTeX gets ugly with all of those fractions. Doesn't it?

LOL! Yes, I did (and yes, it does). How many engineers does it take to solve a simple algebra problem?

Thanks to the edit button, all evidence of our ineptitude has disappeared.