Solving Cubic Equations: Algebraic and Numerical Methods

In this article, we will explore two methods of solving the cubic equation 9x^3 - 9x^2 - 16 0. We will delve into algebraic techniques as well as a numerical approach using the Newton-Raphson algorithm. This comprehensive guide will provide you with a thorough understanding of solving cubic equations, highlighting the advantages of each method.

Algebraic Solution

When solving a cubic equation, such as 9x^3 - 9x^2 - 16 0, one approach involves making substitutions to simplify the equation and transform it into a more manageable form. This method is particularly useful in cases where the coefficients of the cubic equation are not simple integers, or when the equation is derived from a more complex problem.

Step-by-Step Algebraic Solution

Let's consider the original equation:

9x^3 - 9x^2 - 16 0

First Substitution:

We start by substituting x y^{1/3} to eliminate the squared term. This simplifies our equation to:

y^3 - frac{1}{3}y - frac{50}{27} 0

Second Substitution (Vieta's Substitution):

Next, we apply Vieta's substitution y w frac{1}{9w} to transform the equation into a quadratic in terms of w^3. This results in:

w^3 - frac{50}{27}w^3frac{1}{729} 0

This can be rewritten as:

w^3 - frac{50}{27} frac{1}{729w^3} 0

We then multiply through by 729w^3 to get:

729w^6 - 50w^3 1 0

Let z w^3. Then the equation becomes:

z^2 - frac{50}{27}z frac{1}{729} 0

Solving this quadratic equation using the quadratic formula, we obtain:

w^3 frac{25 pm 4sqrt{39}}{27}

Since we are interested in the real root, we take the positive root:

w^3 frac{25 4sqrt{39}}{27}

The next step is to find the cube root of w^3:

w sqrt[3]{frac{25 4sqrt{39}}{27}}

Substituting back, we get:

y w frac{1}{9w}

And reversing the substitutions:

x y^{1/3} left(w frac{1}{9w}right)^{1/3}

The approximate solution for x is:

x approx 1.65167267

Numerical Solution Using Newton-Raphson Algorithm

The second and more practical approach for solving cubic equations involves using a numerical method, such as the Newton-Raphson algorithm. This method is particularly effective for obtaining accurate solutions quickly, especially when dealing with complex or real-world problems.

Newton-Raphson Algorithm

The Newton-Raphson method is an iterative process that refines the estimates of the roots of a function. The basic idea is to use the derivative of the function to approximate the behavior of the function near the current estimate.

Given the function:

y 9x^3 - 9x^2 - 16

We first need to compute the derivative:

y' 27x^2 - 18x

The iterative formula for the Newton-Raphson method is:

x_{n 1} x_n - frac{y_n}{y_n'}

We start with an initial estimate. Since the function is clearly negative for x 1 and positive for x 2, we can begin with x_0 1.5.

Applying the Newton-Raphson method, we perform the iterations:

x_0 1.5

Iteration 1:

x_1 1.5 - frac{9(1.5)^3 - 9(1.5)^2 - 16}{27(1.5)^2 - 18(1.5)} approx 1.64872438

Iteration 2:

x_2 1.64872438 - frac{9(1.64872438)^3 - 9(1.64872438)^2 - 16}{27(1.64872438)^2 - 18(1.64872438)} approx 1.65160254

Iteration 3:

x_3 1.65160254 - frac{9(1.65160254)^3 - 9(1.65160254)^2 - 16}{27(1.65160254)^2 - 18(1.65160254)} approx 1.65167267

After just three iterations, the result is accurate to more than 6 decimal places.

General Form and Substitutions

The general form of a cubic equation is:

x^3 b_1x^2 b_2x b_3 0

To eliminate the x^2-term, we use the substitution x y - frac{b_1}{3}. This substitution results in a reduced cubic equation:

y^3 py q 0

where:

p b_2 - frac{b_1^2}{3}, q frac{2b_1^3}{27} - frac{b_1b_2}{3}b_3

Finally, by letting y w frac{p}{3w}, we transform the equation into a quadratic form:

w^3 frac{p^3}{27w^3} 0

Multiplying through by w^3, we get:

w^6 - frac{p^3}{27} 0

From this, we can find w and then y, and ultimately x.

Conclusion

Both algebraic and numerical methods provide effective ways to solve cubic equations. The algebraic method is useful for understanding the structure of the equation and for finding exact or simplified solutions, while the numerical method is ideal for obtaining precise solutions quickly, especially when iterative refinement is needed.

As an engineer or mathematician, the choice of method often depends on the specific problem at hand and the desired level of accuracy.