Question : If $x^2-xy+y^2=2$ and $x^4+x^2y^2+y^4=6$, then the value of $(x^2+xy+y^2)$ is:
Option 1: 1
Option 2: 12
Option 3: 3
Option 4: 36
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Correct Answer: 3
Solution :
Given: The values of $x^2-xy+y^2=2$ and $x^4+x^2y^2+y^4=6$.
The algebraic identity used is $(x^2+xy+y^2)(x^2-xy+y^2)=x^4+x^2y^2+y^4$
⇒ $(x^2+xy+y^2)(x^2-xy+y^2)=x^4+x^2y^2+y^4$
⇒ $(x^2+xy+y^2)\times2=6$
⇒ $(x^2+xy+y^2)=3$
Hence, the correct answer is 3.
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