If x^{2}+y^{2}+z^{2}=xy+yz+zx,then the value | vedclass

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(D) Given the equation: $x^{2}+y^{2}+z^{2}-xy-yz-zx=0$.Multiplying by $2$ on both sides,we get $2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx=0$.This can be rewrit

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$1$

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(D) Given the equation: $x^{2}+y^{2}+z^{2}-xy-yz-zx=0$.Multiplying by $2$ on both sides,we get $2x^{2}+2y^{2}+2z^{2}-2xy-2yz-2zx=0$.This can be rewritten as $(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=0$.Since the sum of squares is zero,each term must be zero: $x-y=0$,$y-z=0$,and $z-x=0$.This implies $x=y=z$.Let $x=y=z=k$ (where $k 
eq 0$).Substituting $x=y=z=k$ into the expression:$\frac{3k^{4}+7k^{4}+5k^{4}}{5k^{4}+7k^{4}+3k^{4}} = \frac{15k^{4}}{15k^{4}} = 1$.

If $x^{2}+y^{2}+z^{2}=xy+yz+zx$,then the value of $\frac{3x^{4}+7y^{4}+5z^{4}}{5x^{2}y^{2}+7y^{2}z^{2}+3z^{2}x^{2}}$ is

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