If `tanx+tan2x+tan3x=tanxtan2xtan3x` then value of `|sin3x+cos3x|` is ___________

Explanation

To solve the equation \( \tan x + \tan 2x + \tan 3x = \tan x \tan 2x \tan 3x \) and find the value of \( |\sin 3x + \cos 3x| \), we can follow these steps:

### Step 1: Use the property of tangent
We know that if \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \), then \( A + B + C = n\pi \) for some integer \( n \).

In our case, let:
- \( A = x \)
- \( B = 2x \)
- \( C = 3x \)

Thus, we have:
\[ x + 2x + 3x = n\pi \]
This simplifies to:
\[ 6x = n\pi \]

### Step 2: Solve for \( x \)
From the equation \( 6x = n\pi \), we can express \( x \) as:
\[ x = \frac{n\pi}{6} \]

### Step 3: Substitute \( x \) into \( |\sin 3x + \cos 3x| \)
Now we need to find \( |\sin 3x + \cos 3x| \). First, we calculate \( 3x \):
\[ 3x = 3 \left(\frac{n\pi}{6}\right) = \frac{n\pi}{2} \]

### Step 4: Calculate \( \sin 3x \) and \( \cos 3x \)
Now we can find \( \sin 3x \) and \( \cos 3x \):
\[ \sin 3x = \sin\left(\frac{n\pi}{2}\right) \]
\[ \cos 3x = \cos\left(\frac{n\pi}{2}\right) \]

### Step 5: Evaluate \( \sin\left(\frac{n\pi}{2}\right) \) and \( \cos\left(\frac{n\pi}{2}\right) \)
The values of \( \sin\left(\frac{n\pi}{2}\right) \) and \( \cos\left(\frac{n\pi}{2}\right) \) depend on \( n \):
- If \( n \) is even, \( \sin\left(\frac{n\pi}{2}\right) = 0 \) and \( \cos\left(\frac{n\pi}{2}\right) = (-1)^{n/2} \).
- If \( n \) is odd, \( \sin\left(\frac{n\pi}{2}\right) = (-1)^{(n-1)/2} \) and \( \cos\left(\frac{n\pi}{2}\right) = 0 \).

### Step 6: Combine the results
Thus, we can express \( |\sin 3x + \cos 3x| \):
- If \( n \) is even:
\[ |\sin 3x + \cos 3x| = |0 + (-1)^{n/2}| = 1 \]
- If \( n \) is odd:
\[ |\sin 3x + \cos 3x| = |(-1)^{(n-1)/2} + 0| = 1 \]

### Conclusion
In both cases, we find that:
\[ |\sin 3x + \cos 3x| = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]