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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A and B only
Expanding along the first row,
\( \displaystyle \Delta=1(1-x^3)-x(x^2-x^2)+x^2(x^4-x) \)
\( \displaystyle =1-x^3-0+x^6-x^3 \)
\( \displaystyle =x^6-2x^3+1 \)
\( \displaystyle =(1-x^3)^2 \)
Comparing with \( (1+ax^3)^b \) gives a = −1 and b = 2. For the roots of Δ = 0, factorise:
\( \displaystyle 1-x^3=(1-x)(1+x+x^2) \)
\( \displaystyle \Delta=(1-x)^2(1+x+x^2)^2 \)
The factor \( 1+x+x^2 \) has discriminant 1 − 4 < 0, so it has no real root.
So a = −1, b = 2, and the only real root of Δ = 0 is x = 1, a double root.
Checking the statements
(A) \( a=-1 \) and \( b=2 \) — Correct. \( \Delta=(1-x^3)^2 \).
(B) \( x=1 \) is a multiple root of \( \Delta=0 \) — Correct. The factor (1 − x) appears twice.
(C) \( x=1 \) is a simple root of \( \Delta=0 \) — Incorrect. It is a double root, not a simple one.
(D) \( x=3 \) is a simple root of \( \Delta=0 \) — Incorrect. At x = 3, \( \Delta=(1-27)^2=676\ne 0 \).