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Simplify $\sqrt[3]{\frac{9\sqrt{5}}{2\sqrt{3}}\cdot\frac{5\sqrt{2}}{8\sqrt{2}}}$ and rationalize the denominator. The result can be expressed in the form $\frac{\sqrt{a}\sqrt[3]{b}}{c}$, where $a$, $b$, and $c$ are positive integers. What is the minimum possible value of the sum $a+b+c$?

Sagot :

tqiu

Answer:

15+6+2=23

Step-by-step explanation:

[tex]\sqrt[3]{\frac{9\sqrt{5}}{2\sqrt{3}}\cdot\frac{5\sqrt{2}}{8\sqrt{2}}}\\=\sqrt[3]{\frac{9\sqrt{5}}{2\sqrt{3}}\cdot\frac{5}{8}}\\=\sqrt[3]{\frac{45\sqrt{5}}{16\sqrt{3}}}\\=\sqrt[3]{\frac{45\sqrt{15}}{16}}\\=\sqrt[3]{\frac{3\cdot\sqrt{15^3}}{16}}\\=\frac{\sqrt[3]{3}\cdot\sqrt{15}}{2\cdot\sqrt[3]{2}}\\=\frac{\sqrt[3]{6}\cdot\sqrt{15}}{2}[/tex]

we want it in the form of [tex]\frac{\sqrt{a}\sqrt[3]{b}}{c}[/tex], so a = 15, b=6, c=2

also this is the minimum possible value, as we cannot simplify it further

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