a-maths,,3次方程

x^3-3ax^2+8=0有3個不等實根,,試求a值的範圍
let f(x)=x^3-3ax^2+8
f'(x)=3x^2-6ax
Let f'(x)=0
3x^2-6ax=0
x=0 or 2a
f(0)=8
f(2a)=8-4a^3=4(2-a^3)
So if x^3-3ax^2+8=0 has 3 distinct real roots
then it should be
f(0)f(2a)<0
So (2-a^3)<0
a^3>2
a>(2)^1/3
我給條進階題給閣下啦
若 f(x) = x3 - 3kx2 + (k + 3) 在f(x) = 0時有相異三實根，求實數 k 之範圍？