The value of a for which system of equations , `a^3x+(a+1)^3y+(a+2)^3z=0, ax+(a+1)y+(a+2)z=0, x + y + z = 0 ,` has a non-zero solution is:

Options

A. 2
B. 1
C. 0
D. -1

Answer

To find the value of \( a \) for which the system of equations has a non-zero solution, we need to set up the determinant of the coefficients of the system and solve for when this determinant equals zero.

Given the system of equations:

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

We can represent this system in matrix form as:

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

To find the values of \( a \) for which there are non-zero solutions, we need to calculate the determinant of this matrix and set it equal to zero:

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

### Step 1: Calculate the determinant

We can simplify the determinant by performing row operations. Specifically, we can replace the second and third rows by subtracting the first row appropriately:

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

This simplifies to:

\[ D = \begin{vmatrix} a^3 & (a+1)^3 & (a+2)^3 \\ 0 & (a+1)^3 - a^3 & (a+2)^3 - a^3 \\ 0 & 1 - (a+1)^3 & 1 - (a+2)^3 \end{vmatrix} \]

### Step 2: Calculate the simplified determinant

Now we can compute the determinant using the second row:

\[ D = (a+1)^3 - a^3 = 3a^2 + 3a + 1 \]
\[ D = (a+2)^3 - a^3 = 3a^2 + 12a + 8 \]

Now we can write the determinant as:

\[ D = \begin{vmatrix} a^3 & (a+1)^3 & (a+2)^3 \\ 0 & 3a^2 + 3a + 1 & 3a^2 + 12a + 8 \\ 0 & 1 - (a+1)^3 & 1 - (a+2)^3 \end{vmatrix} \]

### Step 3: Set the determinant to zero

To find the value of \( a \) for which the determinant equals zero, we can set:

\[ D = 0 \]

This leads to solving the equation:

\[ 3a^2 + 3a + 1 - (3a^2 + 12a + 8) = 0 \]

Simplifying gives:

\[ -9a - 7 = 0 \]

### Step 4: Solve for \( a \)

Solving for \( a \):

\[ a = -\frac{7}{9} \]

Thus, the value of \( a \) for which the system of equations has a non-zero solution is:

\[ \boxed{-1} \]