f(1)=0, f(2)=29, f(3)=124, f(4)=351.
Let f(x)=ax3+bx2+cx+d.
(1)=f(1)=0⇒a+b+c+d=0;
(2)=f(2)=29⇒8a+4b+2c+d=29;
(3)=f(3)=124⇒27a+9b+3c+d=124;
(4)=f(4)=351⇒64a+16b+4c+d=351.
(5)=(2)-(1)=7a+3b+c=29;
(6)=(3)-(1)=26a+8b+2c=124;
(7)=(4)-(1)=63a+15b+3c=351.
(8)=(6)-2(5)=12a+2b=66⇒6a+b=33;
(9)=(7)-3(5)=42a+6b=264⇒7a+b=44.
(10)=(9)-(8)=a=11; therefore b=33-66=-33; 77-99+c=29, c=51; d=-(a+b+c)=-29.
f(x)=11x3-33x2+51x-29.