Originally Posted by
Prove It It's possible that there's only one tangent that goes through that line and is tangent to the curve, but anyway...
Your tangent is a line, so is of the form y = mx + c.
The gradient of this tangent will be the same as the gradient of the curve at a point. The gradient of the curve at any point is given by the derivative.
So differentiate y = x^3 to get the derivative, and substitute this derivative (as a function of x) as m.
Can you get to there? Can you see what to do next?