Hello, Katykiki!
Divide and simplify: \(\displaystyle \L\,\frac{\sqrt[3]{270x^{20}}}{\sqrt[3]{5x}}\)
Here is a sneaky (but legal) way around the messy part . . .
simplify first!
\(\displaystyle \L\frac{\sqrt[3]{270x^{20}}}{\sqrt[3]{5x}} \:= \:\sqrt[3]{\frac{270x^{20}}{5x}} \:= \:\sqrt[3]{54x^{19}}\)
Then we have: \(\displaystyle \L\,\sqrt[3]{27\,\cdot\,2\,\cdot\,x^{18}\,\cdot\,x} \;= \;\sqrt[3]{27}\,\cdot\,\sqrt[3]{2}\,\cdot\,\sqrt[3]{x^{18}}\,\cdot\,\sqrt[3]{x}\)
\(\displaystyle \L\;\;\;\;= \:3\,\cdot\,\sqrt[3]{2}\,\cdot\, x^6\,\cdot\,\sqrt[3]{x} \:= \:3x^6\cdot\sqrt[3]{2x}\)
Rationalize the denominator of the expression: \(\displaystyle \L\,\frac{\sqrt{6x^8y^9}}{\sqrt{5x^2y^4}}\)
Simplify first . . .
The numerator is: \(\displaystyle \L\,\sqrt{6\cdot x^8\cdot y^8\cdot y} \:=\:\sqrt{6}\cdot\sqrt{x^8}\cdot\sqrt{y^8}\cdot\sqrt{y} \:= \:\sqrt{6}\cdot x^4\cdot y^4\cdot\sqrt{y} \;=\;x^4y^4\sqrt{6y}\)
The denominator is: \(\displaystyle \L\,\sqrt{5\cdot x^2\cdot y^2} \:=\:\sqrt{5}\cdot\sqrt{x^2}\cdot\sqrt{y^4} \;= \;xy^2\sqrt[5]\)
The fraction becomes: \(\displaystyle \L\,\frac{x^4y^4\sqrt{6y}}{xy^2\sqrt{5}} \:= \:\frac{x^3y^2\sqrt{6y}}{\sqrt{5}}\)
Multiply top and bottom by \(\displaystyle \sqrt{5}:\;\;\L\frac{\sqrt{5}}{\sqrt{5}}\,\cdot\,\frac{x^3y^2\sqrt{6y}}{\sqrt{5}} \;=\;\frac{x^3y^2\sqrt{30y}}{5}\)
\(\displaystyle \L\frac{2\,+\,\sqrt[3]{3}}{\sqrt[3]{6}}\)
There is no sneaky way around this one . . .
Multiply top and bottom by \(\displaystyle \sqrt[3]{36}:\;\;\L\frac{\sqrt[3]{36}}{\sqrt[3]{36}}\,\cdot\,\frac{2\,+\,\sqrt[3]{3}}{\sqrt[3]{6}} \:=\:\frac{2\sqrt[3]{36}\,+\,\sqrt[3]{108}}{\sqrt[3]{216}} \;= \;\frac{2\sqrt[3]{36}\,+\,\sqrt[3]{108}}{6}\)
Note that the numerator can be simplified: \(\displaystyle \:\sqrt[3]{108} \:= \:\sqrt[3]{27\,\cdot\,4} \:= \:\sqrt[3]{27}\,\cdot\,\sqrt[3]{4} \:= \:3\sqrt[3]{4}\)
Final answer: \(\displaystyle \L\;\frac{2\sqrt[3]{36}\,+\,3\sqrt[3]{4}}{6}\)
