how do i find the simplest form of the quotient with the given problem of: sqrt3(162)/sqrt3(2)??
+36923 I will assume you mean
\(\sqrt[3]{162}\) /\( \sqrt[3]{2}\)
cubrt(162) / cubrt(2) = cubrt(27 * 6) / cubrt (2) = 3 cubrt(6) / cubrt (2) = 3 cubrt(2)cubrt(3) / cubrt(2) = 3 cubrt (3)
+36923 I will assume you mean
\(\sqrt[3]{162}\) /\( \sqrt[3]{2}\)
cubrt(162) / cubrt(2) = cubrt(27 * 6) / cubrt (2) = 3 cubrt(6) / cubrt (2) = 3 cubrt(2)cubrt(3) / cubrt(2) = 3 cubrt (3)
ElectricPavlov is right. She/he is awesome. Sorry. I don't know if you are a girl or boy.
+1904 The way ElectricPavlov solved the problem is the short way. Here is the long way.
\(\frac{\sqrt[3]{162}}{\sqrt[3]{2}}\)
\(\frac{\sqrt[3]{27\times6}}{\sqrt[3]{2}}\)
\(\frac{\sqrt[3]{27}\sqrt[3]{6}}{\sqrt[3]{2}}\)
\(\frac{3\sqrt[3]{6}}{\sqrt[3]{2}}\)
\(\frac{3\sqrt[3]{6}\sqrt[3]{4}}{\sqrt[3]{2}\sqrt[3]{4}}\)
\(\frac{3\sqrt[3]{6\times4}}{\sqrt[3]{2\times4}}\)
\(\frac{3\sqrt[3]{24}}{\sqrt[3]{8}}\)
\(\frac{3\sqrt[3]{8\times3}}{\sqrt[3]{8}}\)
\(\frac{3\sqrt[3]{8}\sqrt[3]{3}}{\sqrt[3]{8}}\)
\(\frac{3\times2\sqrt[3]{3}}{\sqrt[3]{8}}\)
\(\frac{6\sqrt[3]{3}}{\sqrt[3]{8}}\)
\(\frac{6\sqrt[3]{3}}{2}\)
\(3\sqrt[3]{3}\)
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