Let the polynomial be f(x)=ax3+bx2+cx+d, where a, b, c, d are constants.
Plug in the four points:
f(-2)=2=-8a+4b-2c+d (1)
f(-1)=-1=-a+b-c+d (2)
f(1)=5=a+b+c+d (3)
f(3)=67=27a+9b+3c+d (4)
(3)-(2): 6=2a+2c, 3=a+c so c=3-a (5)
(4)-(3): 62=26a+8b+2c, 31=13a+4b+c=13a+4b+3-a,12a+4b=28, 3a+b=7, b=7-3a (6)
Substitute for a and b in (2): -1=-a+7-3a-3+a+d=-3a+4+d, d=3a-5 (7).
Now we have b, c, d in terms of a. Substitute in (1):
2=-8a+4(7-3a)-2(3-a)+3a-5=-8a+28-12a-6+2a+3a-5=-15a+17, 15a=15, a=1.
Therefore b=7-3=4, c=3-1=2, d=3-5=-2
SOLUTION: (a,b,c,d)=(1,4,2,-2) so f(x)=x3+4x2+2x-2
CHECK:
f(-2)=-8+16-4-2=2 OK
f(-1)=-1+4-2-2=-1 OK
f(1)=1+4+2-2=5 OK
f(3)=27+36+6-2=67 OK
This confirms the polynomial passes through the required points.