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# Difference between revisions of "Cubic polynomial"

A cubic polynomial is a polynomial of the form $ax^3+bx^2+cx+d=0$. A cubic polynomial is a polynomial of degree 3. A univariate cubic polynomial has the form . An equation involving a cubic polynomial is called a cubic equation. A closed-form solution known as the cubic formula exists for the solutions of an arbitrary cubic equation.

## Solving a cubic

If the leading coefficient of the cubic is not 1, then divide both sides by the leading coefficient so it is 1. You get an equation of the form

$x^3 + ax^2 + bx + c = 0$

Now, we will make a change in variables to get rid of the $x^2$ term. If we do the substitution $x = y - \frac{a}{3}$, this does the trick. Our new equation is of the form

$y^3 + py = q$

We do another substitution $y = w - \frac{p}{3w}$, and our new equation is of the form

$w^3 + rw^{-3} = q$

We can now turn this into a quadratic in terms of $w^3$, solve for $w$, and then solve for $y$ and finally $x$. (Here is another way to do it.)

Because cubic polynomials have an odd degree, their end behaviors go in opposite directions, and therefore all cubic polynomials must have at least one real root.

## Problems

Note: The above method will work for any cubic, but the cubic polynomials in these problems have special tricks to solve them.

2013 AIME I Problems/Problem 5 The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.

2014 AIME I Problems/Problem 9 Let $x_1 be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $x_2(x_1+x_3)$.

2010 AIME II Problems/Problem 7 Let $P(z)=z^3+az^2+bz+c$, where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$, $w+9i$, and $2w-4$, where $i^2=-1$. Find $|a+b+c|$.