Question

we have the cube root of 24 over the cube root of 25. how do we rationalize the denominator?

Answer 1

multiply top and bottom by \(\sqrt[3]{5}\)

Answer 2

since \(\sqrt[3]{25}\times \sqrt[3]{5}=\sqrt[3]{5^3}=5\) your denominator is now rationalized

Answer 3

so then what should the answer look like? just =5?

Answer 4

oh no, the denominator is 5

Answer 5

\[\frac{\sqrt[3]{24}}{\sqrt[3]{25}}=\frac{\sqrt[3]{8\times 3}}{\sqrt[3]{5^2}}\times \frac{\sqrt[3]{5}}{\sqrt[3]{5}}=\frac{2\sqrt[3]{15}}{\sqrt[3]{5^3}}=\frac{2\sqrt[3]{15}}{5}\]

Answer 6

i actually did two steps at the same time
\[\sqrt[3]{8}=2\] and also \(3\times 5=15\) for the numerator