If you plot 2b^3 - 9b^2 + 27b - 20, it only has one zero, so don't worry about the three possible solutions. I think Pip got it because when I solve 2b^3 - 9b^2 + 27b - 20 = 0 using the calculator (I'll solve it later by completing the cube, my girlfriend says I'm too obsessed with this problem, she's says she's jealous of the problem because I pay more attention to it than to her), you get b = 1, and when you substitute (7x - 2)^1/3 = 1 you get x=3/7, which is the correct solution. So Pip freaking pwns
EDIT: Welp, I looked up several methods for completing the cube:
http://mathworld.wolfram.com/CubicFormula.html
http://mathforum.org/dr.math/faq/faq...quations2.html
http://mathforum.org/dr.math/faq/faq...equations.html
but I don't really follow any of them. I'll have to further investigate this on my own after finals next week