• Present cubic function and cubic equation.
• Introduce the concept of roots of equal absolute value.
• Show how to predict and calculate equal roots, techniques that will be useful when applied to higher order functions.
• Simplify the depressed cubic.
• Simplify Vieta's substitution.
• Review complex numbers as they apply to a complex cube root.
• Present cubic formula simplified.
• Show that the cubic equation is effectively solved when at least one real root is known.
• Use Newton's Method to calculate one real root.
• Show that the cubic equation can be solved with high-school math.

The simplest cubic function has coefficients ${\displaystyle b=c=0}$, for example:

${\displaystyle y=f(x)=2x^{3}+54}$.

To solve the equation:

${\displaystyle 2x^{3}=-54}$

${\displaystyle x^{3}={\frac {-54}{2}}=-27}$

${\displaystyle x={\sqrt[{3}]{-27}}=-3}$

The function also contains two complex roots that may be found as solutions of the associated quadratic:

${\displaystyle ax^{2}+(ap+b)x+ap^{2}+bp+c=2x^{2}+(-6)x+(18)}$${\displaystyle =x^{2}-3x+9=0}$

${\displaystyle x={\frac {3\pm {\sqrt {9-36}}}{2}}}$${\displaystyle ={\frac {3\pm {\sqrt {-27}}}{2}}}$ ${\displaystyle ={\frac {3\pm {\sqrt {27}}{\sqrt {-1}}}{2}}}$${\displaystyle ={\frac {3\pm (3{\sqrt {3}}){\sqrt {-1}}}{2}}}$

Curve ${\displaystyle f(x)}$ is useful for finding the cube root of a real number.

Solve: ${\displaystyle x={\sqrt[{3}]{N}}.}$

${\displaystyle x^{3}=N}$

${\displaystyle x^{3}-N=0.}$

This is equivalent to finding a root of function ${\displaystyle y=g(x)=x^{3}-N.}$

If you use Newton's method to solve ${\displaystyle g(x),}$ it may be advantageous to put ${\displaystyle N}$ in form ${\displaystyle n(10^{3p})}$ where ${\displaystyle 1\leq n<1000.}$

Then ${\displaystyle {\sqrt[{3}]{N}}={\sqrt[{3}]{n}}(10^{p}).}$ Remember to preserve correct sign of result.

The cubic function ${\displaystyle ax^{3}+bx^{2}+cx+d\ \dots \ (1)}$

Let one value of ${\displaystyle x}$ be ${\displaystyle p+q}$ and another be ${\displaystyle p-q}$.

Substitute these values into the original function in ${\displaystyle x\ (1)}$ and expand.

${\displaystyle +appp+3appq+3apqq+aqqq+bpp+2bpq+bqq+cp+cq+d\ \dots \ (2a)}$

${\displaystyle +appp-3appq+3apqq-aqqq+bpp-2bpq+bqq+cp-cq+d\ \dots \ (3a)}$

${\displaystyle (2a)+(3a),+2appp+6apqq+2bpp+2bqq+2cp+2d\ \dots \ (4)}$

${\displaystyle (2a)-(3a),+6appq+2aqqq+4bpq+2cq\ \dots \ (5)}$

Reduce ${\displaystyle (4)}$ and ${\displaystyle (5)}$ and substitute ${\displaystyle Q}$ for ${\displaystyle qq}$:

${\displaystyle +3Qap+Qb+appp+bpp+cp+d\ \dots \ (4a)}$

${\displaystyle +Qa+3app+2bp+c\ \dots \ (5a)}$

Combine ${\displaystyle (4a)}$ and ${\displaystyle (5a)}$ to eliminate ${\displaystyle Q}$ and produce a function in ${\displaystyle p}$:

${\displaystyle (-8aa)p^{3}+(-8ab)p^{2}+(-2ac-2bb)p+(+ad-bc)\ \dots \ (6)}$

From ${\displaystyle (6),\ c0=ad-bc.}$

If ${\displaystyle c0==0,\ p=0}$ is a solution and function ${\displaystyle (5a)}$ becomes:

${\displaystyle Qa+c=0\ \dots \ (5d)}$

${\displaystyle Q={\frac {-c}{a}}}$

${\displaystyle q={\sqrt {Q}}}$ and two roots of ${\displaystyle (1)}$ are ${\displaystyle 0\pm q}$.

An example

See Figure 2.

${\displaystyle y=x^{3}+2x^{2}-9x-18}$

${\displaystyle c0=ad-bc=1(-18)-2(-9)=0.}$

The function has roots of equal absolute value.

${\displaystyle Q={\frac {-c}{a}}={\frac {9}{1}}=9;\ q={\sqrt {Q}}={\sqrt {9}}=\pm 3}$

The roots of equal absolute value are ${\displaystyle 3,-3}$.

This method works with complex roots of equal absolute value. Consider function: ${\displaystyle y=f(x)=2x^{3}-11x^{2}+98x-539}$ ${\displaystyle a,b,c,d=2,-11,98,-539}$ ${\displaystyle c_{0}=ad-bc=2(-539)-(-11)98=-1078+1078=0.}$ ${\displaystyle f(x)}$ has roots of equal absolute value. ${\displaystyle Q={\frac {-c}{a}}={\frac {-98}{2}}=-49}$ ${\displaystyle q={\sqrt {Q}}={\sqrt {-49}}=\pm 7i}$ Roots of equal absolute value are: ${\displaystyle 7i,-7i.}$

Combine ${\displaystyle 4(a)}$ and ${\displaystyle 5(a)}$ from above to eliminate ${\displaystyle p}$ and produce a function in ${\displaystyle Q}$:

${\displaystyle (64aaaaa)Q^{4}+}$ ${\displaystyle (160aaaac-32aaabb)Q^{3}+}$ ${\displaystyle (132aaacc-56aabbc+4abbbb)Q^{2}+}$ ${\displaystyle (27aaadd-18aabcd+40aaccc+4abbbd-25abbcc+4bbbbc)Q+}$ ${\displaystyle (27aacdd-18abccd+4acccc+4bbbcd-bbccc)\ \dots \ (8)}$

From ${\displaystyle (8)}$ above: ${\displaystyle C0=27aacdd-18abccd+4ac^{4}+4b^{3}cd-bbc^{3}}$ ${\displaystyle =c(27aadd-18abcd+4ac^{3}+4b^{3}d-bbcc)}$

If ${\displaystyle C0==0}$, then ${\displaystyle Q=0}$ is a solution , ${\displaystyle x=p\pm 0=p}$ and ${\displaystyle (4a),(5a)}$ become:

${\displaystyle ax^{3}+bx^{2}+cx+d\ \dots \ (4e)}$

${\displaystyle 3ax^{2}+2bx+c\ \dots \ (5e)}$

If ${\displaystyle C0==0}$ because ${\displaystyle c=0}$, there is a stationary point where ${\displaystyle x=0}$.

${\displaystyle (4e)}$ and ${\displaystyle (5e)}$ can be combined to produce:

${\displaystyle (ac)x^{2}+(-3ad+bc)x+(-2bd+cc)\ \dots \ (4f)}$

${\displaystyle (3a)x^{2}+(2b)x+(c)\ \dots \ (5f)}$

${\displaystyle (4f)}$ and ${\displaystyle (5f)}$ can be combined to produce:

${\displaystyle (-9ad+bc)x+(-6bd+2cc)\ \dots \ (4g)}$

${\displaystyle (+6abd-2acc)x+(-3acd+4bbd-bcc)\ \dots \ (5g)}$

If the original function ${\displaystyle (1)}$ contains 3 unique roots, then ${\displaystyle (4g),(5g)}$ are numerically different.

If the original function ${\displaystyle (1)}$ contains exactly 2 equal roots, then ${\displaystyle (4g),(5g)}$ are numerically identical, and the 2 roots have the value ${\displaystyle x}$ in ${\displaystyle (4g)}$.

If the original function ${\displaystyle (1)}$ contains 3 equal roots, then ${\displaystyle (4g),(5g)}$ are both null, ${\displaystyle (4f),(5f)}$ are numerically identical and ${\displaystyle x={\frac {-b}{3a}}}$.

From equations (4g) and (5g):

${\displaystyle -(+6abd-2acc)(-6bd+2cc)+(-3acd+4bbd-bcc)(-9ad+bc)}$ ${\displaystyle =+27aacdd-18abccd+4acccc+4bbbcd-bbccc}$ ${\displaystyle =C0}$

Examples

No equal roots

Consider function ${\displaystyle y=x^{3}-2x^{2}-5x+6=(x+2)(x-1)(x-3)}$ from ${\displaystyle 4(g),\ x={\frac {-(-9ad+bc)}{(-6bd+2cc)}}={\frac {44}{122}}}$ from ${\displaystyle 5(g),\ x={\frac {-(+6abd-2acc)}{(-3acd+4bbd-bcc)}}={\frac {122}{236}}}$ ${\displaystyle 4(g),5(g)}$ are numerically different.

Exactly 2 equal roots

Consider function ${\displaystyle y=f(x)=x^{3}-3x^{2}-9x-5=(x+1)(x+1)(x-5)}$ from ${\displaystyle 4(g),\ x={\frac {-72}{72}}=-1}$ from ${\displaystyle 5(g),\ x={\frac {72}{-72}}=-1}$ There are 2 equal roots at ${\displaystyle x=-1.}$ See Function_as_product_of_linear_function_and_quadratic above. To calculate all roots: # python code. a,b,c,d = 1,-3,-9,-5 # Associated quadratic: p = -1 A = a B = A*p + b C = B*p + c # Associated linear function: a1 = A b1 = a1*p + B print ('x3 =', -b1/a1) x3 = 5.0 Roots of cubic function ${\displaystyle f(x)=x^{3}-3x^{2}-9x-5}$ are ${\displaystyle -1,-1,5.}$

3 equal roots

Consider function ${\displaystyle y=x^{3}+9x^{2}+27x+27=(x+3)(x+3)(x+3)}$ from ${\displaystyle 4(g),x={\frac {0}{0}}}$ from ${\displaystyle 4(f),\ 27x^{2}+162x+243=x^{2}+6x+9=0}$ from ${\displaystyle 5(f),\ 3x^{2}+18x+27=x^{2}+6x+9=0}$ ${\displaystyle 4(f),5(f)}$ are numerically identical, the discriminant of each is ${\displaystyle 0}$ and ${\displaystyle x={\frac {-6}{2}}=-3}$

The depressed cubic may be used to solve the cubic equation.

In the cubic function: ${\displaystyle y=f(x)=ax^{3}+bx^{2}+cx+d}$ let ${\displaystyle x={\frac {-b+t}{3a}}}$, substitute for ${\displaystyle x}$ and expand:

${\displaystyle y={\frac {t^{3}+(9ac-3bb)t+(27aad-9abc+2b^{3})}{27a^{2}}}}$

When the function is equated to ${\displaystyle 0}$, the depressed equation is:

${\displaystyle t^{3}+At+B=0}$ where

${\displaystyle A=(9ac-3bb)}$ and

${\displaystyle B=(27aad-9abc+2b^{3})}$

In the depressed equation the coefficient of ${\displaystyle t^{3}}$ is ${\displaystyle 1}$ and the coefficient of ${\displaystyle t^{2}}$ is ${\displaystyle 0}$.

The depressed function is a specific case of the general function in which coefficient ${\displaystyle b}$ is missing. Let ${\displaystyle y=ax^{3}+cx+d.}$ Then point of inflection has coordinates ${\displaystyle (0,d),}$ and ${\displaystyle y'=3ax^{2}+c.}$ When ${\displaystyle x==0,\ y'=c.}$ If coefficient ${\displaystyle b}$ is missing, the cubic function becomes ${\displaystyle y=ax^{3}+cx+d,}$ and point of inflection has coordinates ${\displaystyle (0,d)}$ and slope at point of inflection ${\displaystyle =c.}$

Be prepared for the possibility that one or both of ${\displaystyle A,B}$ may be zero.

When A = 0

This condition occurs when the cubic function in ${\displaystyle x}$ has exactly one stationary point or when slope at point of inflection is zero. ${\displaystyle f(x)=4x^{3}-6x^{2}+3x-1}$ ${\displaystyle f(t)=t^{3}-216}$ ${\displaystyle t={\sqrt[{3}]{216}}=6}$ ${\displaystyle x={\frac {-b+t}{3a}}={\frac {6+6}{3(4)}}=1}$ The other roots may be derived from the associated quadratic: ${\displaystyle y=4x^{2}+(4(1)+(-6))x+4(1)(1)+(-6)(1)+3}$${\displaystyle =4x^{2}-2x+1}$ ${\displaystyle x={\frac {2\pm {\sqrt {4-16}}}{8}}}$ ${\displaystyle ={\frac {2\pm {\sqrt {12}}{\sqrt {-1}}}{8}}}$ ${\displaystyle ={\frac {2\pm (2{\sqrt {3}}){\sqrt {-1}}}{8}}}$ ${\displaystyle ={\frac {1\pm {\sqrt {3}}{\sqrt {-1}}}{4}}}$

When B = 0

This condition occurs when the cubic function in ${\displaystyle x}$ is of format ${\displaystyle (x+g)(x+g+h)(x+g-h)}$ or when point of inflection is on the ${\displaystyle X}$ axis. ${\displaystyle f(x)=x^{3}+9x^{2}+31x+39}$ ${\displaystyle f(t)=t^{3}+36t=t(t^{2}+36)}$ ${\displaystyle x={\frac {-b}{3a}}={\frac {-9}{3(1)}}=-3}$ ${\displaystyle t^{2}=-36;\ t={\sqrt {-36}}={\sqrt {36(-1)}}=\pm 6{\sqrt {-1}}}$ ${\displaystyle x={\frac {-b+t}{3a}}={\frac {-9\pm 6{\sqrt {-1}}}{3(1)}}=-3\pm 2{\sqrt {-1}}}$

When A = B = 0

This condition occurs when: slope at point of inflection is ${\displaystyle 0}$, and point of inflection is on ${\displaystyle X}$ axis. Consider function ${\displaystyle y=x^{3}+9x^{2}+27x+27=(x+3)(x+3)(x+3)}$ ${\displaystyle f(t)=t^{3}+(9ac-3bb)t+(27aad-9abc+2b^{3})}$ ${\displaystyle =t^{3}+(9\cdot 1\cdot 27-3\cdot 9\cdot 9)t+(27\cdot 1\cdot 1\cdot 27-9\cdot 1\cdot 9\cdot 27+2\cdot 9\cdot 9\cdot 9)}$ ${\displaystyle =t^{3}+(0)t+(0)}$ ${\displaystyle x={\frac {-b+t}{3a}}={\frac {-b}{3a}}={\frac {-9}{3}}=-3}$

See Vieta's Substitution.

Let the depressed cubic be written as: ${\displaystyle t^{3}-3Ct+B}$ where ${\displaystyle C=b^{2}-3ac}$ and ${\displaystyle A=-3C}$

Let ${\displaystyle t=w+{\frac {C}{w}}={\frac {w^{2}+C}{w}}}$

Substitute for ${\displaystyle t}$ in the depressed function:

${\displaystyle f(w)=w^{6}+Bw^{3}+C^{3}}$