Using properties of determinants, prove the following: ∣∣∣∣1+a1111+b1111+c∣∣∣∣=ab+bc+ca+abc

LHS=Δ= ∣∣∣∣1+a1111+b1111+c∣∣∣∣
Taking out a, b, c common from 1, 2 and 3 row respectively, we get
Δ=abc∣∣∣∣∣∣1a+11a1a1b1b+11b1c1c1c+1∣∣∣∣∣∣
Applying R1→R1+R2+R3
Δ=abc∣∣∣∣∣∣1a+1b+1c+11a+1b+1c+11a+1b+1c+11b1b+11b1c1c1c+1∣∣∣∣∣∣
=abc(1a+1b+1c+1)∣∣∣∣∣1111b1b+11b1c1c1c+1∣∣∣∣∣
Applying C2→C2−C1, C3→C3−C1, we get
Δ=abc(1a+1b+1c+1)∣∣∣∣∣1001b101c01∣∣∣∣∣
=abc(1a+1b+1c+1)×(1×1×1)(∴the determinant of a triangular matrix is the product of its diagonal elements)
=abc(1a+1b+1c+1)=abc(bc+ac+ca+abcabc)=ab+bc+ca+abc= RHS
Hence proved.