"Depressed Cubic" Equations

Scipione del Ferro solved the "Depressed Cubic" equations.

Tartaglia claimed he solved it(But it is just a copyright infringement)

Having a bet with Antonio Fior in solving this type of equations, he took a long time solving all 30 questions.

Problem is that this type of equations are really hard to solve. View the full method in solving "Depressed Cubic" equations online.

I'm going to introduce a more layman way of solving this type of equations.

${ x }^{ 3 }+mx=n$

Factorise out $x$ from left hand side of the equation and find out the factors for n.

Substitute in the factors and make sure both sides of the equation matches. Solve for $x$

Note that this approach only works when the polynomial equation has an integer root. It is a result of (but not equivalent to) the rational root theorem.

#DepressedCubic

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

How do you solve the "simplest" case $x^3 + x = 1$ ?

Your method only works if there is an integer solution.

Calvin Lin Staff - 6 years, 3 months ago

Yes. Only for integers.

Luke Zhang - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$