Question

prove that
${x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx = ( \frac{x - y} { \sqrt{2} })^{2} + ( \frac{y - z}{ \sqrt{2} } )^{2} ( \frac{z - x}{ \sqrt{2} } )^{2}$
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Answer 1

Answer:

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#Here's ur answer..☆☆☆

Step-by-step explanation:

$\boxed{ \huge{ \mathcal{ \pink{answer...}}}}$

To prove...

${x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx = ( \frac{x - y} { \sqrt{2} })^{2} + ( \frac{y - z}{ \sqrt{2} } )^{2} ( \frac{z - x}{ \sqrt{2} } )^{2}$

$\red{ \bold{given....rhs}} \\ = ( \frac{(x - y) }{ \sqrt{2} } ) {}^{2} + ( \frac{(y - z) }{ \sqrt{2} } ) {}^{2} + ( \frac{(z - x)}{ \sqrt{2} } ) {}^{2} \\ = \frac{(x - y) {}^{2} }{2} + \frac{(y - z) {}^{2} }{2} + \frac{(z - x) {}^{2} }{2} \\ = \frac{x {}^{2} - 2xy + {y}^{2} }{2} + \frac{ {y}^{2} - 2yz + {z}^{2} }{2} + \frac{ {z}^{2} - 2zy + {x}^{2} }{2} \\ = \frac{2 {x}^{2} + 2 {y}^{2} + 2 {z}^{2} - 2xy - 2yz - 2zx }{2} \\ = \green{ \bold{ {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx(lhs)}} \\ ...hence \: proved....$

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