What did you do? If what you posted is your answer, then what was the question? If what you posted was the question, what is your answer?humakhan said:do you understand how i did this?
\(\displaystyle \L\frac{\sqrt[3]{x}\,\cdot\,\sqrt{y}\,\cdot\,\left(\sqrt[4]{xy}\right)^3} {x^{-2}\,\cdot\,y^3\,\cdot\,\left(\sqrt{x}\right)^3\,\cdot\,y^{-3}}\)
soroban said:Hello, humakhan!
Are you trying to indicate different roots?
I'll take a guess at what you meant . . .
\(\displaystyle \L\frac{\sqrt[3]{x}\,\cdot\,\sqrt{y}\,\cdot\,\left(\sqrt[4]{xy}\right)^3} {x^{-2}\,\cdot\,y^3\,\cdot\,\left(\sqrt{x}\right)^3\,\cdot\,y^{-3}}\)
First of all, the \(\displaystyle y^3\) and the \(\displaystyle y^{-3}\) in the denominator will cancel.
We have: \(\displaystyle \L\:\frac{x^{\frac{1}{3}}\,\cdot\,y^{\frac{1}{2}}\,\cdot\,\left[\left(xy\right)^{\frac{1}{4}}\right]^3}{x^{-2}\,\cdot\,\left(x^{\frac{1}{2}}\right)^3} \;=\;\frac{x^{\frac{1}{3}}\,\cdot\,y^{\frac{1}{2}}\,\cdot\,(xy)^{\frac{3}{4}}}{x^{-2}\,\cdot\,x^{\frac{3}{2}}} \;=\;\frac{x^{\frac{1}{3}}\,\cdot\,y^{\frac{1}{2}}\,\cdot\,x^{\frac{3}{4}}\,\cdot\,y^{\frac{3}{4}} }{x^{-\frac{1}{2}} }\)
\(\displaystyle \L\;=\;\left(x^{\frac{1}{3}}\,\cdot\,x^{\frac{3}{4}}\,\cdot\,x^{\frac{1}{2}}\right)\,\cdot\,\left(y^{\frac{1}{2}}\,\cdot\,y^{\frac{3}{4}}\right) \;= \;x^{\frac{19}{12}}\,\cdot\,y^{\frac{5}{4}}\)
soroban said:Hello, humakhan!
Are you trying to indicate different roots?
I'll take a guess at what you meant . . .
\(\displaystyle \L\frac{\sqrt[3]{x}\,\cdot\,\sqrt{y}\,\cdot\,\left(\sqrt[4]{xy}\right)^3} {x^{-2}\,\cdot\,y^3\,\cdot\,\left(\sqrt{x}\right)^3\,\cdot\,y^{-3}}\)