=1 non-zero finite So, d=e=f=0 and f(x)=x3(x3+ax2+bx+c) Hence,
lim
x→0
f(x)
x3
=c=1 Now, as f(x)=x6+ax5+bx4+x3 and f′(x)=0 at x=1 and x=−1 i.e. f′(x)=6x5+5ax4+4bx3+3x2 Now, f′(1)=0 ⇒6+5a+4b+3=0 ⇒5a+4b=−9 . . . (i) and f′(−1)=0 ⇒−6+5a−4b+3=0 ⇒5a−4b=3 . . . (ii) From Eqs. (i) and (ii), a=−3∕5 and b=−3∕2 ∴f(x)=x6−