If \(x^4+x^2y^2+y^4=133\) and \(x^2+xy+y^2=19\), then find the value of \(x^2-xy+y^2\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 7
The expression \(x^4+x^2y^2+y^4\) splits into two factors: \(x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)\). One factor is given, so the other one is found by division.
\(x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=(x^2+y^2+xy)(x^2+y^2-xy)\)
\(133=19\times(x^2-xy+y^2)\)
\(x^2-xy+y^2\displaystyle=\frac{133}{19}=7\)