Given the curve:
\[
y = x^3 - 2x + 7
\]
Find the equation of the tangent at the point \( (1, 6) \).
Step 1: Find the derivative \( \frac{dy}{dx} \) which gives the slope of the tangent:
\[
\frac{dy}{dx} = 3x^2 - 2
\]
Step 2: Calculate the slope at \( x = 1 \):
\[
m = 3(1)^2 - 2 = 3 - 2 = 1
\]
Step 3: Use point-slope form of the line:
\[
y - y_1 = m (x - x_1)
\]
where \( (x_1, y_1) = (1, 6) \) and \( m = 1 \):
\[
y - 6 = 1 \times (x - 1)
\]
\[
y - 6 = x - 1
\]
\[
y = x + 5
\]
Therefore, the equation of the tangent is:
\[
\boxed{y = x + 5}
\]