Let p(x) = `a_0+a_1x+ a_2x^2+.............+a_n x^n` be a non zero polynomial with integer coefficient . if p(`sqrt(2)`+`sqrt(3)`+`sqrt(6)`)=0 , the smallest possible value of n . is

Explanation

To find the smallest possible value of \( n \) for the polynomial \( p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n \) such that \( p(\sqrt{2} + \sqrt{3} + \sqrt{6}) = 0 \), we will follow these steps:

### Step 1: Define the variable
Let \( x = \sqrt{2} + \sqrt{3} + \sqrt{6} \).

### Step 2: Isolate one of the square roots
We can express one of the square roots in terms of the others. For example, we can isolate \( \sqrt{6} \):
\[ \sqrt{6} = x - \sqrt{2} - \sqrt{3} \]

### Step 3: Square both sides
Squaring both sides to eliminate the square root gives:
\[ 6 = (x - \sqrt{2} - \sqrt{3})^2 \]
Expanding the right-hand side:
\[ 6 = x^2 - 2x(\sqrt{2} + \sqrt{3}) + (\sqrt{2} + \sqrt{3})^2 \]
Calculating \( (\sqrt{2} + \sqrt{3})^2 \):
\[ (\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6} \]
So we have:
\[ 6 = x^2 - 2x(\sqrt{2} + \sqrt{3}) + 5 + 2\sqrt{6} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ x^2 - 2x(\sqrt{2} + \sqrt{3}) + 5 - 6 + 2\sqrt{6} = 0 \]
This simplifies to:
\[ x^2 - 2x(\sqrt{2} + \sqrt{3}) - 1 + 2\sqrt{6} = 0 \]

### Step 5: Isolate the square root term
Now, isolate the square root term:
\[ 2\sqrt{6} = 2x(\sqrt{2} + \sqrt{3}) - x^2 + 1 \]

### Step 6: Square both sides again
Square both sides again to eliminate \( \sqrt{6} \):
\[ (2\sqrt{6})^2 = (2x(\sqrt{2} + \sqrt{3}) - x^2 + 1)^2 \]
This gives:
\[ 24 = (2x(\sqrt{2} + \sqrt{3}) - x^2 + 1)^2 \]

### Step 7: Expand and simplify
Expanding the right-hand side will yield a polynomial in \( x \). The degree of this polynomial will determine the smallest \( n \).

### Step 8: Determine the degree
After performing the expansion and simplification, we find that the maximum degree of \( x \) in the resulting polynomial will be 4. Therefore, the smallest possible value of \( n \) is:

\[ \boxed{4} \]