Question

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 $a+b+c$?

Answer 1

Compute the product inside the cube root:
(9√5)/(2√3) * (5√2)/(8√2) = (45√5)/(16√3) = (45/16)√(5/3) = (45/16)(√15/3) = (15√15)/16.

So the expression is ∛(15√15/16). Note ∛(15√15) = √15, so
∛(15√15/16) = √15 / ∛16.

Rationalize the cube root in the denominator by multiplying top and bottom by ∛(16^2)=∛256:
√15/∛16 = (√15·∛256)/16.
Since 256 = 2^8 = (2^6)(2^2) we have ∛256 = 4∛4, hence
(√15·∛256)/16 = (√15·4∛4)/16 = (√15·∛4)/4.

Thus one valid representation is (√15·∛4)/4, so (a,b,c)=(15,4,4) and a+b+c = 23.

To see this is minimal, write √a·∛b/c = √15·∛4/4 and note √a must contain the √15 factor, so a = 15s^2 (s∈Z+). Writing √a = s√15 and equating gives ∛b = (c/(4s))∛4, hence b = c^3/(16 s^3). For b to be an integer we must have c = 4fs and then b = 4f^3. So all solutions are
a = 15 s^2, b = 4 f^3, c = 4 f s (f,s∈Z+), and the sum is 15 s^2 + 4 f^3 + 4 f s.
The minimum occurs at f = s = 1, giving the minimum sum 15 + 4 + 4 = 23.

Answer: 23.