| [ taking common (a+b+c) from c1] On applying R2→R2−R1 and R3→R3−R1 =(a+b+c)|
1
b
c
0
c−b
a−c
0
a−b
b−c
| =(a+b+c)[−(b−c)]2−(a−c)(a−b)] =−(a+b+c)[(b−c)2+(a−c)(a−b)] ⇒ Δ=−(a+b+c)(a2+b2+c2−ab−bc−ca) Since, a,b and c are roots of x3+px+q=0 ∴a+b+c=0 Hence, Δ=0