Question : 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:
Option 1: 2
Option 2: 1
Option 3: 0
Option 4: –1
Correct Answer: 1
Solution :
Given: $x^{2}+y^{2}+z^{2}=xy+yz+zx$
⇒ $x^{2}+y^{2}+z^{2}-xy-yz-zx=0$
⇒ $\frac{1}{2}[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}]=0$
So, $(x-y)=0$, $(y-z)=0$, $(z-x)=0$
So, $x=y=z$
Putting this value in the given condition, we have:
$\frac{3x^{4}+7y^{4}+5z^{4}}{5x^{2}y^{2}+7y^{2}z^{2}+3z^{2}x^{2}}=\frac{15x^{4}}{15x^{4}}=1$
Hence, the correct answer is 1.
Related Questions
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Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$?
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