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Question: Answered & Verified by Expert
$\left|\begin{array}{lll}1 & a & a^2-b c \\ 1 & b & b^2-a c \\ 1 & c & c^2-a b\end{array}\right|=$
MathematicsDeterminantsJEE Main
Options:
  • A 0
  • B $a^3+b^3+c^3-3 a b c$
  • C $3 a b c$
  • D $(a+b+c)^3$
Solution:
2602 Upvotes Verified Answer
The correct answer is: 0
$\left|\begin{array}{ccc}1 & a & a^2-b c \\ 1 & b & b^2-a c \\ 1 & c & c^2-a b\end{array}\right|=\left|\begin{array}{ccc}0 & a-b & (a-b)(a+b+c) \\ 0 & b-c & (b-c)(a+b+c) \\ 1 & c & c^2-a b\end{array}\right|$ by $\left\{\begin{array}{l}R_1 \rightarrow R_1-R_2 \\ R_2 \rightarrow R_2-R_3\end{array}\right.$
$(a-b)(b-c)\left|\begin{array}{lll}0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^2-a b\end{array}\right|=0 \quad, \quad\left\{R_1=R_2\right\}$

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