Question

If \(\Delta = \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} = (1 + ax^3)^b\), then which of the following statements are TRUE?
A. \(a = -1\) and \(b = 2\)
B. \(x = 1\) is a multiple root of \(\Delta = 0\)
C. \(x = 1\) is a simple root of \(\Delta = 0\)
D. \(x = 3\) is a simple root of \(\Delta = 0\)
Choose the correct answer from the options given below:

• A. A and D only
• B. A and C only
• C. A, B and D only
• D. A and B only

Hint:

Expand the determinant to get \((1 - x^3)^2\). Then compare with \((1 + ax^3)^b\) and factor \(1 - x^3\).

Answer 1

The Correct Option is D

Step 1: Expand the determinant:
Expand along the first row:
\[ \Delta = 1(1 \cdot 1 - x \cdot x^2) - x(x^2 \cdot 1 - x \cdot x) + x^2(x^2 \cdot x^2 - 1 \cdot x) \] \[ \Delta = (1 - x^3) - x(x^2 - x^2) + x^2(x^4 - x) = 1 - x^3 + x^6 - x^3 \] \[ \Delta = 1 - 2x^3 + x^6 = (1 - x^3)^2 \]

Step 2: Check statement A:
Compare \((1 - x^3)^2\) with \((1 + ax^3)^b\). This gives \(a = -1\) and \(b = 2\). So A is TRUE.

Step 3: Check statements B and C:
\(\Delta = (1-x^3)^2 = (1-x)^2(1 + x + x^2)^2\). The factor \((1-x)\) appears twice, so \(x = 1\) is a root of multiplicity 2, a multiple root. So B is TRUE and C is FALSE.

Step 4: Check statement D:
Put \(x = 3\): \(\Delta = (1 - 27)^2 = 676 \neq 0\). So 3 is not a root at all. D is FALSE.

Final Answer:
Only A and B are true. This is option 4. \[ \boxed{\text{A and B only}} \]