dividing radical expressions

[Katykiki]

so I need help with dividing and simplifying and rationalizing. My teacher went through the steps but didn't explain why to do what she did.

Some example problems are:
dividing and simplify:
[³√(270x^20)]/[ ³√(5x)]

rationalizethe denominator of the expression:
[√(6x^8y^9)]/[√(5x^2y^4)]
or
[2+³√(3)]/ [³√(6)]

I just need someone to work out the steps for me, that would be amazing!

 

[Denis]

[³√(270x^20)]/[ ³√(5x)]

Youtch! That's sure not easy...what grade are you in?

Your expression:
[(270x^20)^(1/3)] / [(5x)^(1/3)]
= [(270x^20)^(1/3)] * [(5x)^(-1/3)] ; rule: a / b^x = a * b^(-x)

Term by term:
270^(1/3) = 27^(1/3) * 10^(1/3) = 3 * 10^(1/3) = 3 * 2^(1/3) * 5^(1/3)

(x^20)^(1/3) = x^(20/3)

(5x)^(-1/3) = 5^(-1/3) * x^(-1/3)

So combining above:
3 * 2^(1/3) * 5^(1/3) * x^(20/3) * 5^(-1/3) * x^(-1/3)

5^(1/3) * 5^(-1/3) = 5^0 = 1 : so those are out

x^(20/3) * x^(-1/3) = x^(19/3)

And we're left with:
3 * 2^(1/3) * x^(19/3)

That is correct (I tested it); however, there may be a better/simpler way;
if there is, I'm sure someone will jump in to humiliate me

NOTE: NO! I ain't doing no more of these for you!

 

[soroban]

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: \(\displaystyle \L\;\frac{2\sqrt[3]{36}\,+\,3\sqrt[3]{4}}{6}\)