If `f(x) = |[x^2-4x+6,2x^2+4x+10,3x^2-2x+16],[x-2,2x+2,3x-1],[1,2,3]|` then find the value of `int_(-3)^3(x^2sinx)/(1+x^6)f(x) dx`.

Answer

To solve the problem, we need to evaluate the integral

\[ \int_{-3}^{3} \frac{x^2 \sin x}{1 + x^6} f(x) \, dx \]

where

\[ f(x) = \left| \begin{array}{ccc} x^2 - 4x + 6 & x - 2 & 1 \\ 2x^2 + 4x + 10 & 2x + 2 & 2 \\ 3x^2 - 2x + 16 & 3x - 1 & 3 \end{array} \right| \]

### Step 1: Simplifying the Determinant

First, we simplify the determinant \( f(x) \). We can factor out constants from the columns of the determinant.

Taking 2 out of the second column, we have:

\[ f(x) = 2 \cdot \left| \begin{array}{ccc} x^2 - 4x + 6 & x - 2 & 1 \\ 2x^2 + 4x + 10 & 1 & 2 \\ 3x^2 - 2x + 16 & 3x - 1 & 3 \end{array} \right| \]

### Step 2: Row Operations

Next, we perform row operations to simplify the determinant further. We can subtract the first column from the second and three times the first column from the third:

\[ f(x) = 2 \cdot \left| \begin{array}{ccc} x^2 - 4x + 6 & x - 2 & 1 \\ 2x^2 + 4x + 10 & 2x + 2 - (x - 2) & 2 \\ 3x^2 - 2x + 16 & 3x - 1 - 3(x - 2) & 3 \end{array} \right| \]

This gives us:

\[ f(x) = 2 \cdot \left| \begin{array}{ccc} x^2 - 4x + 6 & 1 & 1 \\ 2x^2 + 4x + 10 & 6x - 1 & 2 \\ 3x^2 - 2x + 16 & 10x - 2 & 3 \end{array} \right| \]

### Step 3: Expanding the Determinant

Now we can expand the determinant along the third row:

\[ f(x) = 2 \cdot \left( 1 \cdot \left| \begin{array}{cc} 2x^2 + 4x + 10 & 6x - 1 \\ 3x^2 - 2x + 16 & 10x - 2 \end{array} \right| \right) \]

Calculating the determinant of the 2x2 matrix:

\[ = 2 \cdot \left( (2x^2 + 4x + 10)(10x - 2) - (6x - 1)(3x^2 - 2x + 16) \right) \]

### Step 4: Evaluating the Integral

Now we can substitute \( f(x) \) back into the integral:

\[ \int_{-3}^{3} \frac{x^2 \sin x}{1 + x^6} f(x) \, dx = 2 \int_{-3}^{3} \frac{x^2 \sin x}{1 + x^6} \, dx \]

### Step 5: Analyzing the Integrand

Notice that \( x^2 \) and \( 1 + x^6 \) are even functions, while \( \sin x \) is an odd function. The product of an even function and an odd function is an odd function. Therefore, the integrand

\[ \frac{x^2 \sin x}{1 + x^6} \]

is an odd function.

### Step 6: Conclusion

The integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-3}^{3} \frac{x^2 \sin x}{1 + x^6} \, dx = 0 \]

Thus, we conclude:

\[ \int_{-3}^{3} \frac{x^2 \sin x}{1 + x^6} f(x) \, dx = 0 \]

### Final Answer

The value of the integral is:

\[ \boxed{0} \]