To prove:
∣∣
∣∣aa+ba+b+c2a3a+2b4a+3b+2c3a6a+3b10a+6b+3c∣∣
∣∣=a3
Taking L.H.S.
∣∣
∣∣aa+ba+b+c2a3a+2b4a+3b+2c3a6a+3b10a+6b+3c∣∣
∣∣
Applying R2→R2−2R1
=∣∣
∣∣aa+ba+b+c0a2a+b3a6a+3b10a+6b+3c∣∣
∣∣
Applying R3→R3−3R1
=∣∣
∣∣aa+ba+b+c0a2a+b03a7a+3b∣∣
∣∣
=a∣∣∣a2a+b3a7a+3b∣∣∣−0+0
=a(a(7a+3b)−3a(2a+b))
=a(7a2+3ab−6a2−3ab)
=a(a2)
=a3
=R.H.S.
Thus L.H.S.=R.H.S.
Hence proved.