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\[\frac{ \sqrt[6]{c^5d} }{ \sqrt[4]{c^3d^3} }\]

\[c^{m/n}=\sqrt[n]{c^m}\]

The only law that'd help nice would be:\[\large \rm \sqrt[n]{x^m} = x^{m \over n}\]

I know that rule! But then what?

Also, you gotta use:\[\rm {a^x \over a^y} = {a^{x - y}}\]and\[\rm {a^x}\cdot {a^y} = a^{x + y}\]

Then, as JonasK pointed.

\[(\frac{ c^{5/6} }{ c^{3/4} })(\frac{ d^{1/6} }{ d^{3/4} })\]

can you use the rule by @ParthKohli \[\frac{ a^m }{ a^n }=a^{m-n}\]

Jonask, just continue solving please :) All I need is the steps!

\[c^{5/6-3/4}d^{1/6-3/4}\]

same base=subtract powers

cant you just add the numerator with numerator and the denominator with the denominator?

can you illustrate

no we can only addif they are both having same base

cd^5/6+1/6

they have the same base

no c and d are not the same variable \[\frac{ c^n*c^m }{ d^nd^m }=\frac{ c^{m+n} }{ d^{m+n} }\]

Ahh, that's true! Sorry :]

so what is our answer

\[c^{-2/24}d^{-14/12}\] is the answer
lets try this one\[\frac{ x^3y^{-2} }{ \sqrt[3]{x^2y^2}}\]

,,law of exponents" what method of solving is this ?

You convert EVERYTHING to exponents!

yes than this mean very very easy

begin with sqrtx =x^(1/2)

or
a^(3/2)
------- = a^((3/2)-(1/3))
a^(1/3)
right ?

so and i think this will be the law of exponents method sure
yes ?

Yup!

ok
good luck
bye

Thanks! Bye =D