Let's first discuss the case of the equation f(x)=d root about X, D ∈ [-2,2].

When |d|=2, it is determined by (2)

) We know that the two different roots of f(x)=-2 are 1 and12, and we notice that f(x) is odd function.

The two different roots of f (x) = 2 are-1 and 2.

When | d |d| 0, f (1)-d = f (-2)-d =-2-d < 0,

∴ 1 2, - 1, 1, 2

None of them is f (x) = d.

The root of.

From (1), f ′ (x) = 3 (x+1) (x-1).

① when x∈(2, +∞) and f ′ (x) > 0, so f(x) is a monotone increasing function, so f (x) > f (2) = 2.

At this time, f(x)=d has no real root at (2, +∞).

② when x ∈ (1, 2), f ′ (x) > 0, so f(x) is a monotone increasing function.

∵ f (1)-d < 0, f (2)-d > 0, y=f(x)-d is continuous.

∴f(x)=d in (1, 2

) has a unique real root.

Similarly, in (a 2, a I

) has a unique real root.

③ When x∈(- 1, 1), f ′ (x) is less than 0, so f(x) is a monotonically decreasing function.

∵ f (1)-d > 0, f (2)-d < 0, y=f(x)-d is continuous.

∴f(x)=d in (1, 1

) has a unique real root.

Therefore, when |d|=2, f(x)=d has two different roots.

X 1, x2, satisfying |x 1|= 1, | x2 | = 2；; When | d |d|