S saramarie New member Joined Jan 12, 2012 Messages 3 Jan 12, 2012 #1 How to rationalize the numerator? the cubed root of x divided by y
S saramarie New member Joined Jan 12, 2012 Messages 3 Jan 12, 2012 #2 √3: (x)/y I multiplied the top by √3: (x^2) as well as the bottom. but that's as far as i have gotten
√3: (x)/y I multiplied the top by √3: (x^2) as well as the bottom. but that's as far as i have gotten
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jan 12, 2012 #3 Hello, saramarie! How to rationalize the numerator? . . The cubed root of x divided by y Click to expand... \(\displaystyle \text{We have: }\;\dfrac{\sqrt[3]{x}}{y}\) \(\displaystyle \text{Multiply by }\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}: }\; \dfrac{\sqrt[3]{x}}{y}\cdot\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}} \;=\; \dfrac{\sqrt[3]{x^3}}{y\sqrt[3]{x^2}} \;=\; \dfrac{x}{y\sqrt[3]{x^2}} \)
Hello, saramarie! How to rationalize the numerator? . . The cubed root of x divided by y Click to expand... \(\displaystyle \text{We have: }\;\dfrac{\sqrt[3]{x}}{y}\) \(\displaystyle \text{Multiply by }\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}: }\; \dfrac{\sqrt[3]{x}}{y}\cdot\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}} \;=\; \dfrac{\sqrt[3]{x^3}}{y\sqrt[3]{x^2}} \;=\; \dfrac{x}{y\sqrt[3]{x^2}} \)
