\( \left[\begin{array}{c}{(3^{2^{3}})\div (3^{1})^{5}+\sqrt {4^{3}}-39}\end{array}\right]+5^{0}= \)

To solve this expression, let's break it down step by step. First, simplify \( 3^{2^3} \): \[ 2^3 = 8 \] thus, \[ 3^{2^3} = 3^{8} = 6561. \] Next, simplify \( (3^{1})^{5} \): \[ (3^{1})^{5} = 3^{5} = 243. \] Now divide these two results: \[ (3^{2^3}) \div (3^{1})^{5} = 6561 \div 243 = 27. \] This is because \( 6561 \div 243 = 3^{8} \div 3^{5} = 3^{8-5} = 3^{3} = 27. \) Now simplify \( \sqrt{4^{3}} \): \[ 4^{3} = 64 \] thus, \[ \sqrt{4^{3}} = \sqrt{64} = 8. \] Now substitute everything back into the expression: \[ 27 + 8 - 39 + 5^{0}. \] Now calculate \( 5^{0} \): \[ 5^{0} = 1. \] Putting it all together: \[ 27 + 8 - 39 + 1 = 36 - 39 = -3. \] So, the final answer is: \[ -3. \]