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Rationalize the denominator: 1/cube root(a)

Guest Sep 25, 2014

Best Answer 

 #2
avatar+91517 
+5

$$\\\dfrac{1}{\sqrt[3]{a}}\\\\\\
=\dfrac{1}{a^{1/3}}\\\\\\
=\dfrac{1}{a^{1/3}}\times \dfrac{a^{2/3}}{a^{2/3}}\\\\\\
=\dfrac{a^{2/3}}{a^{1/3+2/3}}\\\\\\
=\dfrac{a^{2/3}}{a^{1}}\\\\\\
=\dfrac{\sqrt[3]{a^2}}{a}\\\\\\
$Which is exactly what geno said in the first place! $$$

Melody  Sep 26, 2014
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2+0 Answers

 #1
avatar+17711 
+5

To remove the radical from the denominator, multiply both the numerator and the denominator by a value that makes the denominator whole.

If the denominator is cube root(a) multiply the numerator and denominator by cube root(a²).

The numerator becomes the cube root(a²) while the denominator becomes the cube root(a³), which is just a.

So the answer is: cube root(a²) / a.

geno3141  Sep 25, 2014
 #2
avatar+91517 
+5
Best Answer

$$\\\dfrac{1}{\sqrt[3]{a}}\\\\\\
=\dfrac{1}{a^{1/3}}\\\\\\
=\dfrac{1}{a^{1/3}}\times \dfrac{a^{2/3}}{a^{2/3}}\\\\\\
=\dfrac{a^{2/3}}{a^{1/3+2/3}}\\\\\\
=\dfrac{a^{2/3}}{a^{1}}\\\\\\
=\dfrac{\sqrt[3]{a^2}}{a}\\\\\\
$Which is exactly what geno said in the first place! $$$

Melody  Sep 26, 2014

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