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If $a \neq b \neq c$, the value of $x$ which satisfies the equation$\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0$ is
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Verified Answer
The correct answer is:
$x=0$
Obviously, on putting $x=0$, we observe that the determinant becomes
$\Delta_{x=0}=\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|=a(b c)-b(a c)=0$
$\therefore \quad x=0$ is a root of the given equation.
Aliter : Expanding $\Delta$, we get
$\Delta=-(x-a)[-(x+b)(x-c)]+(x-b)[(x+a)(x+c)]=0$
$\Rightarrow \quad 2 x^3-(2 \Sigma a b) x=0$
$\Rightarrow$ Either $x=0$ or $x^2=\sum a b$ (i.e., $x= \pm \sum a b$ )
Again $x=0$ satisfies the given equation
$\Delta_{x=0}=\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|=a(b c)-b(a c)=0$
$\therefore \quad x=0$ is a root of the given equation.
Aliter : Expanding $\Delta$, we get
$\Delta=-(x-a)[-(x+b)(x-c)]+(x-b)[(x+a)(x+c)]=0$
$\Rightarrow \quad 2 x^3-(2 \Sigma a b) x=0$
$\Rightarrow$ Either $x=0$ or $x^2=\sum a b$ (i.e., $x= \pm \sum a b$ )
Again $x=0$ satisfies the given equation
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