The square root of $x^{2} + y^{2} + z^{2} - 2 x y + 2 y z - 2 z x$

| x + y - z |

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| x - y + z |

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| x + y + z |

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| x - y - z |

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Solution

The correct option is D $| x - y - z |$
$\sqrt{x^{2} + y^{2} + z^{2} - 2 x y + 2 y z - 2 z x}$

$= \sqrt{(x - y - z)^{2}}$ since $[(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 a b + 2 b c - 2 a c]$

$= | x - y - z |$ $[$ mod is applied since $x - y - z$ may be negative $]$

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Similar questions

(x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x)

\frac{x^{2} y z + y^{2} z x + z^{2} x y}{x} = x y z + y^{2} z + z^{2} y

(x y, y z, z x)

x (x - 1) + 2 y z = y (y - 1) + 2 z x = z (z - 1) + 2 x y .

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

x^{2} + y^{2} + z^{2} + x + y + z - 1 = 0

x^{2} + y^{2} + z^{2} + x + y + z - 5 = 0

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