If x+y+z=0, what will be the value of frac{x^ | vedclass

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(C) Given that $x+y+z=0.$Squaring both sides,we get $(x+y+z)^{2}=0^{2}.$Expanding this,we have $x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=0.$Therefore,$x^{2}+y^{2

Vedclass · Accepted Answer

$2$

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(C) Given that $x+y+z=0.$Squaring both sides,we get $(x+y+z)^{2}=0^{2}.$Expanding this,we have $x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=0.$Therefore,$x^{2}+y^{2}+z^{2}=-2(xy+yz+zx).$We can rewrite the expression inside the bracket as $x(y+z)+yz.$Since $x+y+z=0,$ we have $y+z=-x.$Substituting this,we get $x^{2}+y^{2}+z^{2}=-2(x(-x)+yz) = -2(-x^{2}+yz) = 2(x^{2}-yz).$Thus,$\frac{x^{2}+y^{2}+z^{2}}{x^{2}-yz} = \frac{2(x^{2}-yz)}{x^{2}-yz} = 2.$

If $x+y+z=0,$ what will be the value of $\frac{x^{2}+y^{2}+z^{2}}{x^{2}-y z} ?$

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