1. ## My Cubic Solution

If, $\displaystyle b\geq 0$.
Then call,
$\displaystyle x=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}$
Cube,
$\displaystyle x^3=(a+\sqrt{b})+3\sqrt[3]{(a+\sqrt{b})^2}\sqrt[3]{a-\sqrt{b}}+3\sqrt[3]{a+\sqrt{b}}\sqrt{(a-\sqrt{b})^2} +(a-\sqrt{b})$
$\displaystyle x^3=2a+3\sqrt[3]{a-\sqrt{b}}\sqrt[3]{a+\sqrt{b}}\left( \sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)$
$\displaystyle x^3=2a+3\sqrt[3]{a^2-b} \cdot x$
$\displaystyle x^3-3x\sqrt[3]{a^2-b}=2a$.

Example:
Solve,
$\displaystyle x^3+x=1$
We need to find constants $\displaystyle a,b$.
Such that,
$\displaystyle 2a=1$
$\displaystyle -3\sqrt[3]{a^2-b}=1$
Thus,
$\displaystyle a=1/2$
Thus.
$\displaystyle -3\sqrt[3]{(1/2)^2-b}=1$
$\displaystyle \sqrt[3]{1/4-b}=-1/3$
$\displaystyle 1/4-b=-1/27$
$\displaystyle -1/4+b=1/27$
$\displaystyle b=1/27+1/4=29/108\geq 0$
Thus one solution is,
$\displaystyle x=\sqrt[3]{1/2+\sqrt{29/108}}+\sqrt[3]{1/2-\sqrt{29/108}}$

Note, it fails when $\displaystyle b<0$

2. Originally Posted by ThePerfectHacker
If, $\displaystyle b\geq 0$.
Then call,
$\displaystyle x=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}$
Cube,
$\displaystyle x^3=(a+\sqrt{b})+3\sqrt[3]{(a+\sqrt{b})^2}\sqrt[3]{a-\sqrt{b}}+3\sqrt[3]{a+\sqrt{b}}\sqrt{(a-\sqrt{b})^2} +(a-\sqrt{b})$
$\displaystyle x^3=2a+3\sqrt[3]{a-\sqrt{b}}\sqrt[3]{a+\sqrt{b}}\left( \sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}\right)$
$\displaystyle x^3=2a+3\sqrt[3]{a^2-b} \cdot x$
$\displaystyle x^3-3x\sqrt[3]{a^2-b}=2a$.

Example:
Solve,
$\displaystyle x^3+x=1$
We need to find constants $\displaystyle a,b$.
Such that,
$\displaystyle 2a=1$
$\displaystyle -3\sqrt[3]{a^2-b}=1$
Thus,
$\displaystyle a=1/2$
Thus.
$\displaystyle -3\sqrt[3]{(1/2)^2-b}=1$
$\displaystyle \sqrt[3]{1/4-b}=-1/3$
$\displaystyle 1/4-b=-1/27$
$\displaystyle -1/4+b=1/27$
$\displaystyle b=1/27+1/4=29/108\geq 0$
Thus one solution is,
$\displaystyle x=\sqrt[3]{1/2+\sqrt{29/108}}+\sqrt[3]{1/2-\sqrt{29/108}}$

Note, it fails when $\displaystyle b<0$
Hell TPH,

if you did this on your own, then congratulations! But... you are more than 500 years too late. This method is known (for me) as the Cardano formula of the reduced cubic equation. This method was detected first by Scipione del Ferro in 1501 or 1502 and afterwards published by Cardano.

Have a look here: Scipione del Ferro - Wikipedia, the free encyclopedia

Nevertheless you did a great job!

EB

3. Originally Posted by earboth

Nevertheless you did a great job!
I know that, this IS NOT the original solution from the Renaissance. My solution is an alternative quicker techinque for finding one of the zeros.