Answer
The equations of tangent are:$$3x-y-3=0$$ $$3x-y-7=0$$
Work Step by Step
Given curve, $y=x^3-3x^2+3x-3$. The tangent line to this curve is parallel to $3x−y=15$, which means that the slope of the tangent line is $3$ (the slope of the line it is parallel to).
Differentiating y with respect to x,
$$\frac{dy}{dx}=3x^2-6x+3$$This slope must be equal to $3$, that is, the slope of the required tangent.
$$3x^2-6x+3=3\implies3x^2-6x=0$$$$x=0,2$$Thus, the tangent touches the curve y at $(0,y(0))≡(0,-3)$ & $(2,y(2))≡(2,-1)$. Thus, there will be two tangents. For first point,$$y+3=3(x-0)\implies3x-y-3=0.$$For second point, $$y+1=3(x-2)=3x-6$$$$3x-y-7=0.$$
