Let `x=(3sqrt(6)+7)^(89)`. If {x} denotes the fractional part of 'x' then find the remainder when `x{x}+(x{x})^(2)+(x{x})^(3)` is divided by 31.

Answer

To solve the problem, we need to find the remainder when \( x \{x\} + (x \{x\})^2 + (x \{x\})^3 \) is divided by 31, where \( x = (3\sqrt{6} + 7)^{89} \) and \( \{x\} \) denotes the fractional part of \( x \).

### Step 1: Understanding \( x \) and \( \{x\} \)

Let:
\[ x = (3\sqrt{6} + 7)^{89} \]
We can also consider:
\[ y = (3\sqrt{6} - 7)^{89} \]
Since \( 3\sqrt{6} \approx 7.35 \), we have \( 3\sqrt{6} - 7 \) which is a small positive number. Thus, \( y \) will be a very small positive number as \( 89 \) is a large exponent.

### Step 2: Finding \( \{x\} \)

The fractional part \( \{x\} \) can be expressed as:
\[ \{x\} = x - \lfloor x \rfloor \]
Using the binomial theorem, we can write:
\[ x + y = (3\sqrt{6} + 7)^{89} + (3\sqrt{6} - 7)^{89} \]
This sum will be an integer because all terms involving odd powers of \( 3\sqrt{6} \) will cancel out, leaving only even powers which are integers.

Thus, we can express:
\[ \lfloor x \rfloor = (3\sqrt{6} + 7)^{89} + (3\sqrt{6} - 7)^{89} - y \]
This implies:
\[ \{x\} = (3\sqrt{6} + 7)^{89} - \left( (3\sqrt{6} + 7)^{89} + (3\sqrt{6} - 7)^{89} - y \right) \]
Simplifying gives:
\[ \{x\} = y = (3\sqrt{6} - 7)^{89} \]

### Step 3: Calculate \( x \{x\} \)

Now we compute:
\[ x \{x\} = x \cdot y = (3\sqrt{6} + 7)^{89} \cdot (3\sqrt{6} - 7)^{89} \]
Using the difference of squares:
\[ x \{x\} = ((3\sqrt{6})^2 - 7^2)^{89} = (54 - 49)^{89} = 5^{89} \]

### Step 4: Calculate \( x \{x\} + (x \{x\})^2 + (x \{x\})^3 \)

Now we need to compute:
\[ x \{x\} + (x \{x\})^2 + (x \{x\})^3 = 5^{89} + (5^{89})^2 + (5^{89})^3 \]
This can be factored as:
\[ 5^{89} (1 + 5^{89} + (5^{89})^2) \]

### Step 5: Simplifying the expression

Let \( z = 5^{89} \). Then:
\[ z(1 + z + z^2) \]
The polynomial \( 1 + z + z^2 \) can be evaluated modulo 31.

### Step 6: Finding \( 5^{89} \mod 31 \)

Using Fermat's Little Theorem:
\[ 5^{30} \equiv 1 \mod 31 \]
Thus:
\[ 5^{89} = 5^{30 \cdot 2 + 29} \equiv 5^{29} \mod 31 \]

Now we compute \( 5^{29} \mod 31 \):
- \( 5^1 \equiv 5 \)
- \( 5^2 \equiv 25 \)
- \( 5^3 \equiv 125 \equiv 1 \mod 31 \)

Since \( 5^3 \equiv 1 \), we can reduce \( 5^{29} \):
\[ 5^{29} = (5^3)^9 \cdot 5^2 \equiv 1^9 \cdot 25 \equiv 25 \mod 31 \]

### Step 7: Evaluating \( 1 + 5^{89} + (5^{89})^2 \)

Now substituting back:
\[ 1 + 5^{29} + (5^{29})^2 \equiv 1 + 25 + 25^2 \mod 31 \]
Calculating \( 25^2 \mod 31 \):
\[ 25^2 = 625 \equiv 5 \mod 31 \]
Thus:
\[ 1 + 25 + 5 \equiv 31 \equiv 0 \mod 31 \]

### Step 8: Final Calculation

Now we have:
\[ 5^{89} (1 + 5^{89} + (5^{89})^2) \equiv 25 \cdot 0 \equiv 0 \mod 31 \]

### Conclusion

The remainder when \( x \{x\} + (x \{x\})^2 + (x \{x\})^3 \) is divided by 31 is:
\[ \boxed{0} \]