I'm not sure if i'm right, but

expand:

(x^2+n)(x-n)

=x^3-x^2*n+nx-n^2

take the first derivative

so f'(x)=3x^2-2xn+n

-f'(x)=3x^2-2xn+n

m=-(3*(-x)^2-2*-x*n+n)

m=-3x^2-2xn-n

so the tangent line is

y=mx+b

y=(-3x^2-2xn-n)*x+b

b depends on your n, i don't feel like working it out though sorry

I think that's right but i've never done a problem like this so...