Ta có : f(x) = ax3 + 4x(x2-x) - 4x + 8
= ax3 + 4x3 - 4x2 - 4x + 11 - 3
= x3 (a + 4) - 4x(x + 1) + 11-3
f(x) = g (x) ⇔⇔ x3 (a + 4) - 4x(x + 1) +11-3 = x3 - 4x(bx + 1) + c-3
⇔⇔ ⎧⎩⎨⎪⎪a+4=1x+1=bx+1c=11{a+4=1x+1=bx+1c=11 ⎧⎩⎨⎪⎪ a=−3b=1c=11
vậy a = -3 , b = 1 và c = 11