Most math students can solve linear equations -- equations that contain a variable such as "x" without exponents -- with little trouble. Solving quadratic equations -- equations in which the variable is raised to the power of two, such as "x^2" -- is a little more complex. However, solving cubic equations -- equations with a "x^3" term -- requires many more steps and poses problems to even those extremely adept at algebra. This difficulty can be attributed to the shape of a cubic equation, which can look similar to a roller coaster track. You can follow these steps in a linear manner, and with practice you will be able to quickly solve cubic equations.
Write the cubic equation in standard form ax^3 + bx^2 + cx + d = 0. For example, if the equation you wish to solve is x^3 = 7x + 6, rewrite it as x^3 – 7x – 6 = 0.
Find one of the roots by substitution methods. Use trial and error by plugging in values for “x” until one root is found. Call this root “r1.” In the previous example, we can try x = 1, which fails, and then try x = -1, which results in 1^3 – 7(1) – 6 = 0, which holds true. Now you know one root, r1 = -1.
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Use the factor theorem to rewrite the equation. Factor (x – r1) out of the equation. You will be left with (x – r1)(x^2 + ax + b) = 0. In the example, you will rewrite the equation as (x + 1)(x^2 + ax + b) = 0.
Apply synthetic division to the original cubic equation to yield a quadratic expression. Write the resulting quadratic expression as x^2 + dx + f. Applying the process of synthetic division to the original cubic equation in the example yields x^2 – x – 6.
Multiply the first root factor and the quadratic expression together and set it equal to zero. In short, you will have the equation (x – r1)(x^2 + dx +f). For the example, the equation is (x + 1)(x^2 – x – 6) = 0.
Factor this new equation. Since the first root factor is already factored, you technically only need to factor the quadratic expression. You will yield an equation of the form (x – r1)(x – r2)(x – r3) = 0. In the example, the result is (x + 1)(x – 3)(x + 2) = 0.
Find the roots of this equation. These roots are the solutions to the original cubic equation. The roots are simply the numbers you see on the left-hand side of the equation, each multiplied by -1. Hence, the solutions for “x” are “r1,” “r2” and “r3.” In the example, the solutions are x = -1, x = 3 and x = -2.