Maths - Cubic Functions

A cubic function has an x³ term, its general form is:

a x³ + b x² + c x + d = 0

The three solutions are:

root1 = -a/3 + ((-2a³ + 9ab - 27c + sqrt((2a³ -9ab +27c)² + 4(-a² + 3b)³))/54)^1/3 + ((-2a³ + 9ab - 27c - √((2a³ -9ab +27c)² + 4(-a² + 3b)³))/54)^1/3

root2 = -a/3 - (1 + i sqrt(3))*0.5*((-2a³ + 9ab - 27c + sqrt((2a³ -9ab +27c)² + 4(-a² + 3b)³))/54)^1/3 + (-1 + i sqrt(3))*0.5*((-2a³ + 9ab - 27c - sqrt((2a³ -9ab +27c)² + 4(-a² + 3b)³))/54)^1/3

root3 = -a/3 + ((-2a³ + 9ab - 27c + sqrt((2a³ -9ab +27c)² + 4(-a² + 3b)³))/54)^1/3 + ((-2a³ + 9ab - 27c - √((2a³ -9ab +27c)² + 4(-a² + 3b)³))/54)^1/3

Methods of solution

1. by factors
2. by completing the cobe
3. by formulae
4. by graph

Sum and difference of two cubes

(a + b)(a² - ab + b²)

= a (a² - ab + b²) + b (a² - ab + b²)

= a³ - a²b + ab² + ba² - ab² + b³

= a³ + b³

simarly

(a - b)(a² + ab + b²) = a³ - b³

Using Symmetry

Instead of using the equation:

a x³ + b x² + c x + d = 0

= 0

we can use the equation:

(x - x₁)*(a' x² + b' x + c') = 0

where x is our variable and x₁, a', b' and c' are constants.

which has solutions at:

x = x₁ and

In other words, can we use this to reduce the cubic to a quadratic in this way? First we have to multiply out the terms to see if its a general cubic:

(x - x₁)*(a' x² + b' x + c') = (a' x³ + b' x² + c' x ) - x₁*a' x² - x₁* b' x - x₁* c')

= a' x³ + (b' - x₁*a')x² + (c' - x₁* b') x - x₁* c')

Therefore:

• a = a'
• b = b' - x₁*a
• c = c' - x₁* b'
• d = - x₁* c'

So, puting this in terms of the original equation,

• a' = a
• b' = b + x₁*a = b - d*a/c'
• c' = c + x₁* b' = c - d*(b - d*a)/c'
• x₁ = -d/c'

so:

c'² - c*c' + d*(b - d*a) = 0

c' = (c± √(c² - 4*d*(b - d*a)))/2

substituting in above gives:

• a' = a
• b' = b + x₁*a = b - d*a/(c± √(c² - 4*d*(b - d*a)))/2
• c' = (c± √(c² - 4*d*(b - d*a)))/2
• x₁ = -d/(c± √(c² - 4*d*(b - d*a)))/2

Code

Here is a Java function to return the cubic roots

/**
This is incomplete and does not handle roots of -ve, let me know if you can help

code to solve cubic equation ax^3 + bx^2 + cx + d =0
result we be returned in resultReal[0],resultImaginary[0]
to resultReal[2] and resultImaginary[2]
expects arrays to be setup before being called, like this:
double[] resultReal= new double[3];
double[] resultImaginary = new double[3];
*/
public void cubic(double a, double b, double c, double d,
     double[] resultReal , double[] resultImaginary) {
  if (Math.abs(a) < 0.000001) { // if a=0 equation is quadratic
    // handle seperately to avoid divide by zero
	quadratic(b,c,d,resultReal,resultImaginary);
	// the quadratic function is defined on this page:
    // http://www.euclideanspace.com/maths/algebra/equations/polynomial/quadratic/
	return;
  }
  // recuring terms
  double t1=(2*a*a*a -9*a*b +27*c);
  double t2=(-a*a + 3*b);
  double t3= t1*t1 + 4*t2*t2*t2;
  if (t3<0) handle complex roots;
  double t4=((-t1 + sqrt(t3))/54);
  double t41=((-t1 - sqrt(t3))/54);
  double t5=(-a*a + 3b);
  double t6=sqrt(t1*t1 + 4*t5*t5*t5);
  double t7=((-t1 - t6)/54);
  resultReal[0] = -a/3 + Math.pow(t4,1/3) + Math.pow(t7,1/3);
  resultReal[1] = -a/3 - 0.5*Math.pow(t4,1/3)+  -0.5*Math.pow(t4,1/3);
  resultReal[2] = -a/3 + Math.pow(t4,1/3) + Math.pow(t41,1/3)
  resultImaginary[0] = 0;
  resultImaginary[1] = sqrt(t3)*0.5*Math.pow(t4,1/3)+ sqrt(t3)*0.5*Math.pow(t4,1/3);
  resultImaginary[2] = 0;
  return;
} 

Square Root of Minus One

Symmetry