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Question

Find all the solutions to [3]15x - 1 + [3]13x + 1 = 4 [3]x.Enter all the solutions, separated by commas.

Accepted Answer

From the given equation, [3]15x - 1 + [3]13x + 1 - 4 [3]x = 0.We can also write this as [3]15x - 1 + [3]13x + 1 + [3]-64x = 0.Let a = [3]15x - 1, b = [3]13x + 1, and c = [3]-64x, so a + b + c = 0. From the factorization a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ab - bc),we have that a^3 + b^3 + c^3 = 3abc. Hence, -36x = 3 [3](15x - 1)(13x + 1)(-64x).We can simplify this to 3x = [3](15x - 1)(13x + 1)x.Cubing both sides we, get 27x^3 = 195x^3 + 2x^2 - x, so 168x^3 + 2x^2 - x = 0. This factors as x(14x - 1)(12x + 1) = 0, so the solutions are 0, 114, -112.