Let us take the LHS of the given expression.
LHS=∣∣
∣∣abca−bb−cc−ab+cc+aa+b∣∣
∣∣
Perform C1→C1+C2+C3LHS=∣∣
∣∣a+b+cbc0b−cc−a2(a+b+c)c+aa+b∣∣
∣∣
Take (a+b+c) common from C1,
LHS=(a+b+c)∣∣
∣∣1bc0b−cc−a2c+aa+b∣∣
∣∣
Perform R3→R3−2R1LHS=(a+b+c)∣∣
∣∣1bc0b−cc−a0c+a−2ba+b−2c∣∣
∣∣
Now, expanding the determinant along C1, we get
LHS=(a+b+c)((b−c)(a+b−2c)−(c+a−2b)(c−a))
=(a+b+c)(ab+b2−2bc−ac−bc+2c2−c2+ac−ac+a2+2bc−2ab)
=(a+b+c)(a2+b2+c2−ab−bc−ac)
We know that
a3+b3+c3=(a+b+c)(a2+b2+c2−ab−bc−ac)+3abc
⇒a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)
∴LHS=a3+b3+c3−3abc
Thus, LHS = RHS