If \( f(x)=\begin{vmatrix}1 & x & x^2\\x & x^2 & 1\\x^2 & 1 & x\end{vmatrix} \), then the value of \( f(\sqrt[3]{3}) \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( -4 \)
Expanding the determinant along the first row:
\( f(x)=1(x^3-1)-x(x^2-x^2)+x^2(x-x^4) \)
\( =x^3-1+x^3-x^6 \)
\( =-(x^6-2x^3+1) \)
\( =-(x^3-1)^2 \)
At \( x=\sqrt[3]{3} \), \( x^3=3 \):
\( f(\sqrt[3]{3})=-(3-1)^2 \)
\( =-4 \)
Why the other options are incorrect
Option (1) \( -6 \) — Incorrect. The value is −4, not −6.
Option (2) 6 — Incorrect. The value is −4, not 6.
Option (3) 4 — Incorrect. The expansion gives the negative of a perfect square, so the value cannot be positive.