What is \left \lfloor ( 3 + \sqrt{5} ) ^3 \right \rfloor?

(3+√5)^3 = 143.55

so, take it from there.

To find the value of \(\left \lfloor (3 + \sqrt{5})^3 \right \rfloor\), we need to follow these steps:

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, \(\left \lfloor (3 + \sqrt{5})^3 \right \rfloor = 72 + 32\sqrt{5}\).