Question 64

If $$x^4+x^2y^2+y^4=21$$ and $$x^2+xy+y^2=3$$, then what is the value of $$\left(-xy\right)$$?

Solution

$$x^4+x^2y^2+y^4=21$$......(1)

$$x^2+xy+y^2=3$$

$$x^2+y^2=3-xy$$

$$\left(x^2+y^2\right)^2=\left(3-xy\right)^2$$

$$x^4+y^4+2x^2y^2=9+x^2y^2-6xy$$

$$x^4+y^4+x^2y^2=9-6xy$$

$$21=9-6xy$$ [From (1)]

$$-6xy=12$$

$$-xy=2$$

Hence, the correct answer is Option B

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