I am not getting the same answer as the book for this question. it is to find the equation of the normal line at the given point

y = 2x^(3) + x^(-3) where x = -1

I solved for y getting 2 giving the point (-1,2)

taking the derivative I got y' = 6x^2 -3x^-4 and for f'(-1) I got 3

giving me the equation

y-2=3(x+1)

so for the slope of the normal line I take the negative reciprocal being -1/3

so I have

y-2 = [-1/3][x+1] I multiply everything by 3 to get [3y-6] = -1[x+1]

which gives me

x + 3y + 5 = 0

but the book has the answer down as being x+3y + 10 = 0

thank you!