x3+ax2+bx+c=0=(x−a)(x−b)(x−c) a+b+c=−a ⇒2a+b+c=0 ...(i) ab+bc+ca=b...(ii) abc=−c⇒ab=−1[∵c≠0] ...(iii) Also a is a root of equation ⇒2a3+ab+c=0⇒2a3−1+c=0 ⇒c=1−2a3 from (1) 2a2+ab+ac=0 2a2−1+a(1−2a3)=0 2a2−2a4+a−1=0 2a2(1−a)(1+a)+(a−1)=0 ⇒(1−a)[2a2(a+1)−1]=0 ⇒a=1 or 2a3+2a2−1=0 when a=1,b=
−1
a
=−1 and c=1−2a3=−1 when 2a3+2a2−1=0 There will be only one real solution of f(x)=2x3+2x2−1=0 as f′(x)=6x2+4x=0⇒x=0,
−2
3
f(0).f(
−2
3
)<0 ∴ corresponding to this real value of a one triplet is possible ∴ Exactly two triplets (a,b,c) are possible