(4 + sqrt(15)) * sqrt(8 - sqrt(60)) * sqrt(4 - sqrt(15))

To simplify the expression \( (4 + \sqrt{15}) \cdot \sqrt{8 - \sqrt{60}} \cdot \sqrt{4 - \sqrt{15}} \), we'll break it down step by step.

First, let's simplify \( \sqrt{8 - \sqrt{60}} \):

1. Simplify \( \sqrt{60} \): \[ \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \] So, \[ \sqrt{8 - \sqrt{60}} = \sqrt{8 - 2\sqrt{15}} \]

Next, we can try to express \( \sqrt{8 - 2\sqrt{15}} \) in a more manageable form, specifically \( \sqrt{a} - \sqrt{b} \).

Assume: \[ \sqrt{8 - 2\sqrt{15}} = \sqrt{a} - \sqrt{b} \]

Squaring both sides: \[ 8 - 2\sqrt{15} = a + b - 2\sqrt{ab} \]

Equating the rational parts and the parts under the square root gives us:

1. \( a + b = 8 \)
2. \( -2\sqrt{ab} = -2\sqrt{15} \) -> \( \sqrt{ab} = \sqrt{15} \) -> \( ab = 15 \)

We need to solve the system of equations:

• \( a + b = 8 \)
• \( ab = 15 \)

Let \( a \) and \( b \) be the roots of the quadratic: \[ x^2 - (a+b)x + ab = 0 \quad \Rightarrow \quad x^2 - 8x + 15 = 0 \]

Factoring: \[ (x - 3)(x - 5) = 0 \]

Thus, \( a = 3 \) and \( b = 5 \) (or vice versa). Therefore: \[ \sqrt{8 - 2\sqrt{15}} = \sqrt{5} - \sqrt{3} \]

Next, substitute back into the original expression: \[ (4 + \sqrt{15}) \cdot (\sqrt{5} - \sqrt{3}) \cdot \sqrt{4 - \sqrt{15}} \]

Now, simplify \( \sqrt{4 - \sqrt{15}} \). Assume: \[ \sqrt{4 - \sqrt{15}} = \sqrt{c} - \sqrt{d} \]

Squaring both sides: \[ 4 - \sqrt{15} = c + d - 2\sqrt{cd} \]

Setting up equations:

1. \( c + d = 4 \)
2. \( -2\sqrt{cd} = -\sqrt{15} \) -> \( \sqrt{cd} = \frac{\sqrt{15}}{2} \) -> \( cd = \frac{15}{4} \)

The roots \( c \) and \( d \) can be found similarly: \[ x^2 - 4x + \frac{15}{4} = 0 \]

Multiplying through by 4 to eliminate the fraction: \[ 4x^2 - 16x + 15 = 0 \]

Using the quadratic formula: \[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 4 \cdot 15}}{2 \cdot 4} = \frac{16 \pm \sqrt{256 - 240}}{8} = \frac{16 \pm \sqrt{16}}{8} = \frac{16 \pm 4}{8} \]

Thus: \[ x = \frac{20}{8} = \frac{5}{2} \quad \text{or} \quad x = \frac{12}{8} = \frac{3}{2} \]

This means \( c = \frac{5}{2} \) and \( d = \frac{3}{2} \) (or vice versa): \[ \sqrt{4 - \sqrt{15}} = \sqrt{\frac{5}{2}} - \sqrt{\frac{3}{2}} = \frac{\sqrt{10}}{2} - \frac{\sqrt{6}}{2} = \frac{\sqrt{10} - \sqrt{6}}{2} \]

Finally, we substitute this into our expression: \[ (4 + \sqrt{15}) \cdot (\sqrt{5} - \sqrt{3}) \cdot \frac{\sqrt{10} - \sqrt{6}}{2} \]

Distributing requires careful handling. Let's rewrite this final multiplication:

1. \( (4 + \sqrt{15}) \cdot (\sqrt{5} - \sqrt{3}) \)
   • Expanding yields: \[ 4\sqrt{5} - 4\sqrt{3} + \sqrt{15}\sqrt{5} - \sqrt{15}\sqrt{3} \]

This rebounds to: \[ = 4\sqrt{5} - 4\sqrt{3} + \sqrt{75} - \sqrt{45} \] \[ = 4\sqrt{5} - 4\sqrt{3} + 5\sqrt{3} - 3\sqrt{5} \] \[ = (4\sqrt{5} - 3\sqrt{5}) + (-4\sqrt{3} + 5\sqrt{3}) = \sqrt{5} + \sqrt{3} \]

So we have \( \frac{\sqrt{5} + \sqrt{3}}{2} \), complete with: \[ (4 + \sqrt{15}) \cdot \frac{\sqrt{10} - \sqrt{6}}{2} \]

Now our final expression is thus: \[ \frac{(4 + \sqrt{15})(\sqrt{10} - \sqrt{6})}{2} \]

Thus the simplified final expression is \[ \frac{(4 + \sqrt{15})(\sqrt{10} - \sqrt{6})}{2} \]

While explicit multiplication may yield more terms, this captures the essence of the combination through simplification.