it is shown in calculus slope of line tangent to y=x^3 at point (a,a^3) has slope3a^2
find equation of line tangent to y=x^3 at (2,8) and (a, a^3)
At $\displaystyle (2,8)$, the slope of the tangent line is $\displaystyle m=3a^2=3(2)^2=12$. Now use your point-slope form, $\displaystyle y-y_0 = m\left(x-x_0\right)$ to form the equation of the tangent line. The second part can be done the same way.