Find the equation of the tangent to the curve `y=(x^3-1)(x-2)` at the points where the curve cuts the x-axis.

Explanation

To find the equation of the tangent to the curve \( y = (x^3 - 1)(x - 2) \) at the points where it cuts the x-axis, we will follow these steps:

### Step 1: Find the points where the curve cuts the x-axis
The curve cuts the x-axis where \( y = 0 \). Therefore, we set the equation to zero:

\[ (x^3 - 1)(x - 2) = 0 \]

This gives us two factors to solve:

1. \( x^3 - 1 = 0 \)
2. \( x - 2 = 0 \)

From \( x - 2 = 0 \), we get:

\[ x = 2 \]

From \( x^3 - 1 = 0 \), we can factor it as:

\[ x^3 - 1 = (x - 1)(x^2 + x + 1) = 0 \]

The real root from this factorization is:

\[ x = 1 \]

The quadratic \( x^2 + x + 1 \) has no real roots (its discriminant is negative). Therefore, the curve cuts the x-axis at \( x = 1 \) and \( x = 2 \).

### Step 2: Identify the points on the curve
The points where the curve intersects the x-axis are:

1. \( (1, 0) \)
2. \( (2, 0) \)

### Step 3: Differentiate the curve to find the slope
To find the slope of the tangent at these points, we need to differentiate \( y \):

\[ y = (x^3 - 1)(x - 2) \]

Using the product rule:

\[ \frac{dy}{dx} = (x^3 - 1) \cdot \frac{d}{dx}(x - 2) + (x - 2) \cdot \frac{d}{dx}(x^3 - 1) \]

Calculating the derivatives:

\[ \frac{d}{dx}(x - 2) = 1 \]
\[ \frac{d}{dx}(x^3 - 1) = 3x^2 \]

Substituting back into the derivative:

\[ \frac{dy}{dx} = (x^3 - 1)(1) + (x - 2)(3x^2) \]
\[ = x^3 - 1 + 3x^2(x - 2) \]
\[ = x^3 - 1 + 3x^3 - 6x^2 \]
\[ = 4x^3 - 6x^2 - 1 \]

### Step 4: Calculate the slope at the points \( (1, 0) \) and \( (2, 0) \)
1. **At \( (1, 0) \)**:

\[ \frac{dy}{dx} \bigg|_{x=1} = 4(1)^3 - 6(1)^2 - 1 = 4 - 6 - 1 = -3 \]

2. **At \( (2, 0) \)**:

\[ \frac{dy}{dx} \bigg|_{x=2} = 4(2)^3 - 6(2)^2 - 1 = 32 - 24 - 1 = 7 \]

### Step 5: Write the equations of the tangents
Using the point-slope form of the line \( y - y_1 = m(x - x_1) \):

1. **For the point \( (1, 0) \)** with slope \( -3 \):

\[ y - 0 = -3(x - 1) \]
\[ y = -3x + 3 \]

2. **For the point \( (2, 0) \)** with slope \( 7 \):

\[ y - 0 = 7(x - 2) \]
\[ y = 7x - 14 \]

### Final Answer
The equations of the tangents to the curve at the points where it cuts the x-axis are:

1. \( y = -3x + 3 \)
2. \( y = 7x - 14 \)