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Suppose $P (x) = |\begin{array}{ccc} x & - 51 & - 71 \\ 51 & x & - 73 \\ 71 & 73 & x \end{array}|$ .Product of zeros of $P (x)$ is

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Correct option is A

TipSuppose P(x)=x3+ax2+bx+c then product of zero of P(x) is −c=−P(0)Now, −c=−P(0)=−0−51−71510−7371730=0[∴Skew symmetric determinant of odd order]

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Suppose P(x)=x−51−7151x−737173x .Product of zeros of P(x) is

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Suppose $P (x) = |\begin{array}{ccc} x & - 51 & - 71 \\ 51 & x & - 73 \\ 71 & 73 & x \end{array}|$ .Product of zeros of $P (x)$ is