ϕ1(x)=x+a1,ϕ2(x)=x2+b1x+b2,x1=2,x2=3,x3=5Δ=∣∣
∣
∣∣111ϕ1(x1)ϕ1(x2)ϕ1(x3)ϕ2(x1)ϕ2(x2)ϕ2(x3)∣∣
∣
∣∣applyingC3→C3−C2andC2→C2−C1weobtainΔ=∣∣
∣
∣∣100ϕ1(x1)x2−x1x3−x2ϕ2(x1)x22−x21+b1(x2−x1)x23−x22+b(x3−x2)∣∣
∣
∣∣=(x2−x1)(x3−x2)∣∣∣11x2+x1+b1x3+x2+b1∣∣∣=(x2−x1)(x3−x2)(x3−x1)∴Δ=6