** A. Demanet **devised an interesting method of solution of trinomial equations which depends on the use of communicating vessels of convenient forms.

To solve an equation of the third degree of the form:

*x*^{3}+*x*=*c*where

**is a constant, an inverted cone and a cylinder, joined together by means of a tube, are taken.**

*c*As shown below.

The radius * r* of the cone and its height

**are in the ratio:**

*h*

*r*/*h*= √3/√*π*while the base of the cylinder is taken as

**1 cm**

^{2}If ** c** cubic centimeters of water are poured into one of the two vessels, the water will rise to the same height

**in both. The volume of water contained in the cone will be**

*h*

**h****, that in the cylinder**

^{3}**, so that we get:**

*h*

**h**

^{3}+*h*=*c*Therefore, by measuring the height

**of the water we obtain a solution of the equation.**

*h*In the case of the equation

**x**^{3} – *x* = *c*

the cone alone is used, and a solid cylindrical piece whose base is 1 cm^{2} is introduced. The volume ** c** of water poured in will thus be the difference between

**h****and h, and therefore**

^{3}**, the height of the liquid, is again a solution.**

*h*By a substitution **z**** = x√p** we can reduce all reducible equations of the third degree such as:

**x**

^{3}+*pz*=*q*where

**and**

*p***are given positive numbers, to the form**

*q*

**x**

^{3}+*x*=*c***Source: page 253 of “Matematica dilettevole e curiosa” by I. Ghersi*

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