∵y=x^3-1
∴y'=3x²
则y'(1)=3即为点(1,0)处的切线的斜率
∴切线方程为：y-0=3(x-1),即3x-y-3=0
又∵法线的斜率为-1/3
∴法线方程为：y-0=-1/3*(x-1),即x+3y-1=0