What will come in place of both the question marks (?) in the following question ?
$$\frac{(?)^{\frac{4}{3}}}{32}=\frac{128}{(?)^{\frac{5}{3}}}$$
Let the missing number be y
$$\frac{(y)^{\frac{4}{3}}}{32}=\frac{128}{(y)^{\frac{5}{3}}}$$
$$(y)^(\frac{4}{3} + \frac{5}{3})$$ = $$(2)^{12}$$
y = $$(2)^4$$
y = 16
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