If a line makes angles `alpha, beta, gamma, delta` with the diafonals of a cubes then the value of `9(cos 2alpha + cos 2beta + cos2gamma` `+ cos 2 delta )^(2)` equals ….......

Answer

To solve the problem, we need to find the value of \( 9(\cos 2\alpha + \cos 2\beta + \cos 2\gamma + \cos 2\delta)^2 \).

### Step 1: Use the identity for cosine
We know that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Using this identity, we can rewrite each cosine term:
\[ \cos 2\alpha = 2\cos^2 \alpha - 1 \]
\[ \cos 2\beta = 2\cos^2 \beta - 1 \]
\[ \cos 2\gamma = 2\cos^2 \gamma - 1 \]
\[ \cos 2\delta = 2\cos^2 \delta - 1 \]

### Step 2: Substitute into the expression
Now substituting these into the expression:
\[ \cos 2\alpha + \cos 2\beta + \cos 2\gamma + \cos 2\delta = (2\cos^2 \alpha - 1) + (2\cos^2 \beta - 1) + (2\cos^2 \gamma - 1) + (2\cos^2 \delta - 1) \]
This simplifies to:
\[ = 2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta) - 4 \]

### Step 3: Use the property of cosines
From the problem, we know that:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{4}{3} \]
Substituting this into our expression gives:
\[ = 2\left(\frac{4}{3}\right) - 4 = \frac{8}{3} - 4 = \frac{8}{3} - \frac{12}{3} = -\frac{4}{3} \]

### Step 4: Final calculation
Now substituting back into our original expression:
\[ 9\left(-\frac{4}{3}\right)^2 = 9 \cdot \frac{16}{9} = 16 \]

### Conclusion
Thus, the final value is:
\[ \boxed{16} \]