How do you find the common tangent line between: $ y = {x^3} - 3x + 4\;{\text{and}}\;y = 3({x^2} - x)? $

Hint: First differentiate both the curves with respect to $ x $ in order to get the equation of tangents. Now solve the equation of tangents to get their intersection point, that where the tangents intersect. After finding the intersection point find the slope of the tangent after by putting the ordinate of the intersection point in the first derivative of any given equation. Now you have the slope and a point of the tangent line, write its equation in slope point form.

Complete step-by-step answer:
To find the common tangent line between $ y = {x^3} - 3x + 4\;{\text{and}}\;y = 3({x^2} - x) $ we will first differentiate them and find the common point of their tangents by solving them,
 $
   \Rightarrow y = {x^3} - 3x + 4\;{\text{and}}\;y = 3({x^2} - x) \\
   \Rightarrow y' = 3{x^2} - 3\;{\text{and}}\;y' = 6x - 3 \;
  $
Putting $ y' = 3{x^2} - 3 $ in the second equation, we will get
 $
   \Rightarrow 3{x^2} - 3 = 6x - 3 \\
   \Rightarrow 3{x^2} - 6x = 0 \\
   \Rightarrow 3x(x - 2) = 0 \\
   \Rightarrow x = 0\;{\text{and}}\;x = 2 \;
  $
Now we will find the ordinate of the coordinate by putting the x-values into the given curve equations as follows
\[
   \Rightarrow y = {0^3} - 3 \times 0 + 4\;{\text{and}}\;y = {2^3} - 3 \times 2 + 4 \\
   \Rightarrow y = 0 - 0 + 4\;{\text{and}}\;y = 8 - 6 + 4 \\
   \Rightarrow y = 4\;{\text{and}}\;y = 6 \;
 \]
So we get the coordinates of common tangent of the curves: $ (0,\;4)\;{\text{and}}\;(2,\;6) $
Checking this coordinates in the second curve’s equation, if they also are satisfying the curve or not
 $
   \Rightarrow y = 3({0^2} - 0)\;{\text{and}}\;y = 3({2^2} - 2) \\
   \Rightarrow y = 3(0 - 0)\;{\text{and}}\;y = 3(4 - 2) \\
   \Rightarrow y = 0\;{\text{and}}\;y = 6 \;
  $
The first coordinate $ (0,\;4) $ is not satisfying the second curve’s equation, so we will take $ (2,\;6) $ as the common coordinate of tangents.
Now putting $ x = 2 $ in first derivative of the curves equation to get slope of the tangent
 $
   \Rightarrow y' = 3 \times {2^2} - 3\;{\text{and}}\;y' = 6 \times 2 - 3 \\
   \Rightarrow y' = 12 - 3\;{\text{and}}\;y' = 12 - 3 \\
   \Rightarrow y' = 9\;{\text{and}}\;y' = 9 \;
  $
That is slope $ = 9 $
Writing the equation of tangent in slope point form with slope $ = 9 $ and coordinate of point $ \equiv (2,\;6) $
 $
  (y - 6) = 9(x - 2) \\
   \Rightarrow y = 9x - 18 + 6 \\
   \Rightarrow y = 9x - 12 \;
  $
So, the correct answer is “y = 9x - 12”.

Note: When you find points of tangent or a given curve then make sure that you checked the point by substituting its values in the given equation of curve. Sometimes you get multiple coordinates (as we have got in this question) in which few of them don't satisfy the given equations.