Question : If $a^2+b^2+c^2=a b+b c+a c$, then the value of $\frac{11 a^4+13 b^4+17 c^4}{17 a^2 b^2+9 b^2 c^2+15 c^2 a^2}$ is:
Option 1: 1
Option 2: 2
Option 3: 11
Option 4: 4
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Correct Answer: 1
Solution :
Given: $a^2+b^2+c^2=a b+b c+a c$
Multiplying both sides with 2, we get:
$2a^2+2b^2+2c^2=2ab+2bc+2ac$
⇒ $a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac=0$
⇒ $(a-b)^2+(b-c)^2+(c-a)^2=0$
So, $a-b=0$, $b-c=0$, $c-a=0$
⇒ $a=b=c$
Now, $\frac{11 a^4+13 b^4+17 c^4}{17 a^2 b^2+9 b^2 c^2+15 c^2 a^2}$
$=\frac{11 a^4+13 a^4+17 a^4}{17 a^2 a^2+9 a^2 a^2+15 a^2 a^2}$
$=\frac{41 a^4}{41 a^4}$
$=1$
Hence, the correct answer is 1.
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