Find the point on the curve y=x³-x+5 at which the equation of the tangent is y=x

Let the required point of contact be (x, y)
Given: The equation of the curve is y=x³-x+5
The equation of the tangent to the circle as y=x (which is of the form y = mx + c)
Therefore, Slope of the tangent = 1
Now, the slope of the tangent to the given circle at the point (x, y) is given by dy/dx=3x²-1
Then, we have:
3x²-1=1
3x²=2
x²=2/3
x=±√(2/3)
When x=√(2/3), y=(√(2/3))³-√(2/3)+5 = 2√(2/3)/3 - √(2/3) + 5 = -√(2/3)/3 + 5
When x=-√(2/3), y=(-√(2/3))³+√(2/3)+5 = √(2/3)/3 + √(2/3) + 5 = 4√(2/3)/3 + 5
So, the required points are (√(2/3), -√(2/3)/3 + 5) and (-√(2/3), 4√(2/3)/3 + 5)