If  a,b,c∈R and the equations ax2+bx+c=0 and x3+3x2+3x+2=0 have two roots in common, then

We have,x3+3x2+3x+2=0⇒(x+1)3+1=0⇒(x+1+1)[(x+1)2−(x+1)+1]=0⇒(x+2)(x2+x+1)=0⇒x=−2,−1±3i2⇒x=−2,ω,ω2Since, a,b,c∈R,ax2+bx+c=0 cannot have one real and one imaginary root. Therefore, two common roots of ax2+bx+c=0 and x3+3x2+3x+2=0 are ω,ω2thus,−ba=ω+ω2=−1⇒a=band ca=ω⋅ω2=1⇒c=a⇒a=b=c