Hi. I posted this in the Homework question but after 152 views with no right answer, my question looks analytical and rigorous enough to be posted here. Thank you.... ---- I am asking about part iv). [PLAIN]http://img715.imageshack.us/img715/7977/113ivb.jpg [Broken] Attempt at a solution In the given fact, I think [itex] x^3 - x - m(x - a) [/itex] distance from the cubic function to the line. So for every point in [itex] (b, c) [/itex], this would be negative. I'm really not sure how to show [itex] c = -2b [/itex] so I just tried to play with some algebra... At x = b... [itex] b^3 - b - m(b - a) = 0 [/itex] At x = c... [itex] c^3 - c - m(c - a) = 0 [/itex] So they're both equal to 0... [itex] b^3 - b - m(b - a) = c^3 - c - m(c - a) [/itex] so [itex] b^3 - b - mb = c^3 - c - mc [/itex] so by part i) [itex] b^3 - b - b(3b^2 - 1) = c^3 - c - c(3b^2 - 1) [/itex] so [itex] b^3 - b - 3b^3 + b = c^3 - c - 3b^2c + c[/itex] so [itex] -2b^3 = c^3 - 3b^2c [/itex] so [itex] b^2(3c - 2b) = c^3 [/itex] but this doesn't look useful... Thank you.