If x + y + z = 0 and xaybzcyczaxbzbxcya = kabcbcacab then k =

x+y+z=0xaybzcyczaxbzbxcya = kabcbcacab ⇒ xyz a3+b3+c3-3abc=-ka3+b3+c3-3abc∴ k=-xyz

If x + y + z = 0 and xaybzcyczaxbzbxcya = kabcbcacab then k =

x+y+z=0xaybzcyczaxbzbxcya = kabcbcacab ⇒ xyz a3+b3+c3-3abc=-ka3+b3+c3-3abc∴ k=-xyz

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If x + y + z = 0 and $|\begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix}| = k |\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}|$ then k =

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x + y + z = 0

xy + yz + zx

-xyz

$x^{2} + y^{2} + z^{2}$

answer is C.

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$x + y + z = 0$

$|\begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix}| = k |\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}|$

$\begin{array}{rcl} \Rightarrow & xyz (a^{3} + b^{3} + c^{3} - 3 abc) = - k (a^{3} + b^{3} + c^{3} - 3 abc) \\ ∴ k & = & - xyz \end{array}$

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If x + y + z = 0 and xaybzcyczaxbzbxcya = kabcbcacab then k =

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If x + y + z = 0 and $|\begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix}| = k |\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}|$ then k =