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if x y z are all distinct and vmatrix x x 2 1 x 3
Question:
If x,y,z are all distinct and
\(\begin{vmatrix} x & x^2 & 1+x^3\\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix}\)
=0, then the value of xyz is
CUET (PG) - 2023
CUET (PG)
Updated On:
July 22, 2025
-2
-1
-3
0
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The Correct Option is
B
Solution and Explanation
The correct answer is(B): -1
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