calculus!URGENT
posted by Allison on .
Show that the tangent line to the curve y=x^3 at the point x=a also hits the curve at the point x=2a.
Any help?! PLEASE!

dy/dx = 3x^2
at x=a , y = x^3 and dy/dx = 3a^2
equation of tangent:
y = 3a^2x + b
but (a,a^3) is on it, so
a^3 = 3a^2(a) + b
b = 2a^3
tangent equation: y = (3a^2)x  2a^3
intersect that line with y = x^3
x^3 = 3a^2x  2a^3
x^3  3a^2x + 2a^3 = 0
We already know that xa is a factor, since x=a is a solution.
so it factors to
(xa)(x^2 + ax  2a^2) = 0
(xa)(xa)(x  2a) = 0
so x=a, or x=a, or x = 2a 
second line should have been 
at x=a , y = x^3 and dy/dx = 3a^2 
thank you soooooo much!