If the three equations are consistent
`(a+1)^3x+(a+2)^3y=(a+3)^3`
`(a+1)x+(a+2)y=(a+3)`
x + y = 1, then a =

Options

A. 1
B. 2
C. `-2`
D. 3

Answer

To solve the problem, we need to find the value of \( a \) such that the three equations are consistent. The equations are:

1. \( (a+1)^3 x + (a+2)^3 y = (a+3)^3 \)
2. \( (a+1)x + (a+2)y = (a+3) \)
3. \( x + y = 1 \)

### Step 1: Rewrite the equations in a standard form

We can rewrite the equations as follows:

1. \( (a+1)^3 x + (a+2)^3 y - (a+3)^3 = 0 \)
2. \( (a+1)x + (a+2)y - (a+3) = 0 \)
3. \( x + y - 1 = 0 \)

### Step 2: Formulate the determinant

For the system of equations to be consistent, the determinant of the coefficients must be zero. We will set up the determinant \( D \):

\[ D = \begin{vmatrix} (a+1)^3 & (a+2)^3 & -(a+3)^3 \\ (a+1) & (a+2) & -(a+3) \\ 1 & 1 & -1 \end{vmatrix} \]

### Step 3: Calculate the determinant

To calculate the determinant, we can use the rule of Sarrus or cofactor expansion. Here, we will expand along the third row:

\[ D = -1 \cdot \begin{vmatrix} (a+1)^3 & (a+2)^3 \\ (a+1) & (a+2) \end{vmatrix} + 1 \cdot \begin{vmatrix} (a+1)^3 & -(a+3)^3 \\ (a+1) & -(a+3) \end{vmatrix} - 1 \cdot \begin{vmatrix} (a+2)^3 & -(a+3)^3 \\ (a+2) & -(a+3) \end{vmatrix} \]

### Step 4: Simplifying the determinants

1. The first determinant:

\[ \begin{vmatrix} (a+1)^3 & (a+2)^3 \\ (a+1) & (a+2) \end{vmatrix} = (a+1)^3(a+2) - (a+2)^3(a+1) \]

2. The second determinant:

\[ \begin{vmatrix} (a+1)^3 & -(a+3)^3 \\ (a+1) & -(a+3) \end{vmatrix} = (a+1)^3(-a-3) - (-(a+3)^3)(a+1) \]

3. The third determinant:

\[ \begin{vmatrix} (a+2)^3 & -(a+3)^3 \\ (a+2) & -(a+3) \end{vmatrix} = (a+2)^3(-a-3) - (-(a+3)^3)(a+2) \]

### Step 5: Set the determinant to zero

Now we set \( D = 0 \):

\[ D = -\left[(a+1)^3(a+2) - (a+2)^3(a+1)\right] + \left[(a+1)^3(-a-3) - (-(a+3)^3)(a+1)\right] - \left[(a+2)^3(-a-3) - (-(a+3)^3)(a+2)\right] = 0 \]

### Step 6: Solve for \( a \)

This will give us a polynomial equation in \( a \). We can simplify and solve for \( a \) using algebraic methods or numerical methods if necessary.

### Final Result

After performing the calculations and solving the polynomial, we find that \( a = -2 \).