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由 thepiano »
第 2 題
f(x - 1) = x^3 - x^2 + x - 1 = x^2(x - 1) + (x - 1) = (x - 1)(x^2 + 1) = (x - 1)[(x - 1)^2 + 2(x - 1) + 2]
故 f(x) = x(x^2 + 2x + 2)
也可這樣做
令 y = x - 1,x = y + 1
f(y) = (y + 1)^3 - (y + 1)^2 + y + 1 - 1 = (y + 1)^2(y + 1 - 1) + y = y[(y + 1)^2 + 1] = y(y^2 + 2y + 2)
故 f(x) = x(x^2 + 2x + 2)