(3+√5)^3 = 143.55
so, take it from there.
so, take it from there.
Step 1: Evaluate \((3 + \sqrt{5})^3\).
Step 2: Take the floor of the result from step 1.
Let's break it down step by step:
Step 1: Evaluate \((3 + \sqrt{5})^3\)
To find the result of \((3 + \sqrt{5})^3\), we follow the standard order of operations.
\((3 + \sqrt{5})^3\) can be expanded using the binomial theorem:
\((3 + \sqrt{5})^3 = \binom{3}{0} (3)^3 (\sqrt{5})^0 + \binom{3}{1} (3)^2 (\sqrt{5})^1 + \binom{3}{2} (3)^1 (\sqrt{5})^2 + \binom{3}{3} (3)^0 (\sqrt{5})^3\)
Simplifying each term:
\(\binom{3}{0} (3)^3 (\sqrt{5})^0 = 1 \cdot 3^3 \cdot 1 = 27\)
\(\binom{3}{1} (3)^2 (\sqrt{5})^1 = 3 \cdot 3^2 \cdot \sqrt{5} = 27\sqrt{5}\)
\(\binom{3}{2} (3)^1 (\sqrt{5})^2 = 3 \cdot 3^1 \cdot 5 = 45\)
\(\binom{3}{3} (3)^0 (\sqrt{5})^3 = 1 \cdot 3^0 \cdot 5\sqrt{5} = 5\sqrt{5}\)
Adding all the terms together:
\((3 + \sqrt{5})^3 = 27 + 27\sqrt{5} + 45 + 5\sqrt{5} = 72 + 32\sqrt{5}\)
So, we have evaluated \((3 + \sqrt{5})^3\) as \(72 + 32\sqrt{5}\).
Step 2: Take the floor of the result from step 1.
To take the floor of a number, we drop the decimal part and keep only the integer part. Since \(72 + 32\sqrt{5}\) is already an integer, we don't need to make any changes.
Therefore, \(\left \lfloor (3 + \sqrt{5})^3 \right \rfloor = 72 + 32\sqrt{5}\).