Let `ax^(3) + bx^(2) + cx + d`
`|(x+1,2x,3x),(2x+3,x+1,x),(2-x,3x+4,5x-1)|`, then
What is the value of a + b + c + d ?

Options

A. 62
B. 63
C. 65
D. 68

Answer

To solve the problem, we need to evaluate the determinant given by the matrix:

\[ \begin{vmatrix} x + 1 & 2x & 3x \\ 2x + 3 & x + 1 & x \\ 2 - x & 3x + 4 & 5x - 1 \end{vmatrix} \]

and find the coefficients \(a\), \(b\), \(c\), and \(d\) of the polynomial \(ax^3 + bx^2 + cx + d\) such that the determinant is equal to this polynomial.

### Step 1: Substitute \(x = 1\)

We will first substitute \(x = 1\) into the determinant to simplify our calculations.

\[ \begin{vmatrix} 1 + 1 & 2(1) & 3(1) \\ 2(1) + 3 & 1 + 1 & 1 \\ 2 - 1 & 3(1) + 4 & 5(1) - 1 \end{vmatrix} = \begin{vmatrix} 2 & 2 & 3 \\ 5 & 2 & 1 \\ 1 & 7 & 4 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Now we will calculate the determinant using the formula for a 3x3 matrix:

\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where the matrix is:

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix:

- \(a = 2\), \(b = 2\), \(c = 3\)
- \(d = 5\), \(e = 2\), \(f = 1\)
- \(g = 1\), \(h = 7\), \(i = 4\)

Calculating the determinant:

\[ \text{Det} = 2(2 \cdot 4 - 1 \cdot 7) - 2(5 \cdot 4 - 1 \cdot 1) + 3(5 \cdot 7 - 2 \cdot 1) \]

Calculating each term:

1. \(2(8 - 7) = 2(1) = 2\)
2. \(-2(20 - 1) = -2(19) = -38\)
3. \(3(35 - 2) = 3(33) = 99\)

Putting it all together:

\[ \text{Det} = 2 - 38 + 99 = 63 \]

### Step 3: Identify the Polynomial

The determinant evaluates to \(63\), which corresponds to the polynomial \(ax^3 + bx^2 + cx + d\). Since we evaluated the determinant at \(x = 1\), we can say:

\[ a(1^3) + b(1^2) + c(1) + d = 63 \]

This simplifies to:

\[ a + b + c + d = 63 \]

### Final Answer

Thus, the value of \(a + b + c + d\) is:

\[ \boxed{63} \]