+0

Rationalize the denominator: 1/cube root(a)

0
630
2

Rationalize the denominator: 1/cube root(a)

Guest Sep 25, 2014

#2
+94105
+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 #1 +17746 +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 +94105 +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