Let f'(x)=0: x = √ 2.

f(√2)=5-4√2，f(-√2)=5+4√2

So when a∈(5-4√2, 5+4√2), the equation f(x)=a about x has three different real roots.

When x> is at 1, f(x)≥k(x- 1)

f(x)=x^3-6x+5=(x- 1)(x？ +x-5)

Because x-1>; 0, so k≤f(x)/(x- 1)=x? +x-5

g(x)=x？ +x-5 increases monotonically at (1, +∞), so k≤g( 1)=-4.