Question : $\text { If } x^2+y^2+z^2=x y+y z+z x \text { and } x=1 \text {, then find the value of } \frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$.
Option 1: 2
Option 2: 0
Option 3: –1
Option 4: 1
Correct Answer: 1
Solution :
$x^2+y^2+z^2=x y+y z+z x$
$⇒\frac{1}{2}[(x-y)^2+(y-z)^2+(z-x)^2] = 0$
$⇒ x= y, y=z,$ and $z=x$
As $x =1$, so $y=z=1$
So, $\frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$
$= \frac{10(1)^4+5(1)^4+7(1)^4}{13(1)^2(1)^2+6(1)^2(1)^2+3(1)^2(1)^2}$
$=\frac{22}{22}$
$=1$
Hence, the correct answer is 1.
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