Evaluate ∣∣ ∣∣xyx+yyx+yxx+yxy∣∣ ∣∣
Let A=∣∣ ∣∣xyx+yyx+yxx+yxy∣∣ ∣∣
=∣∣
∣
∣∣2(x+y)yx+y2(x+y)x+yx2(x+y)xy∣∣
∣
∣∣
(using C1→C1+C2+C3)
=2(x+y)∣∣
∣∣1yx+y1x+yx1xy∣∣
∣∣ |Taking out 2(x+y) common from C1|
=2(x+y)∣∣
∣∣1yx+y0x−y0x−y−x∣∣
∣∣
(using R2→R2−R1;R3→R3−R1)
Expanding along R1, we get
=2(x+y)×1(−x2+y(x−y))=2(x+y)(−x2+xy−y2)=−2(x+y)(x2−xy+y2)=−2(x3+y3)