Geometrical solution of the cubic equation  

This method originally comes from Cardanos Ars Magna

 

 

This is in modern notation the solution of an equation of the reduced form; x3+mx=n, where m and n are positive numbers because all the dimension in the picture must be positive. 

 

 

First imagine a cube:

(the original picture in Cardanos Ars Magna)

 

 

The length of Cube B at the right-front-bottom is u.
The length of Cube A at the left-back-top is t-u.
Imagine a prism below Cube A [u(t-u)2],
Imagine a prism on top of Cube B [u2(t-u)],

Except all the solids mentioned above, there are two others.
Each of them is tu(t-u) ,

The length of Main Cube is t. Its volume equals the total volume of the six solids,
therefore, t3=u3+(t-u)3+2tu(t-u)+(t-u)u2+u(t-u)2

Re arrange: t3-u3=(t-u)3+[2tu(t-u)+(t-u)u2+u(t-u)2]

t3-u3=(t-u)3+(t-u)[2tu+u2+u(t-u)]

t3-u3=(t-u)3+(t-u)(3tu)

Now, if we let x=t-u then, t3-u3=x3+3tux
Compare this with x3+mx=n, we get

m=3tu and n=t3-u3

Combine the two statements above to get:

Now we have to solve a
quadrated equation in t 3

 

Hence we get the positive root:

But we know that u3=t3-n, so:

And finally the solution becomes:

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