By Andryan A. A.
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Extra info for A Boundary Value Problem in a Strip for Partial Differential Equations in Classes of Tempered Functions
All the sequential triplicates constitute fifteen types, and they can be solved using the properties of the circle. The triplicate ones in every four consecutive ranks constitute twenty-four types, and can be solved using conic sections. Further, all the quartiles among every four consecutive ranks constitute twenty-eight types, and can be solved using properties of conic sections. So all the types (which fall within these seven ranks that can be solved using our methods) amount to eighty-six; only six were mentioned in the books of ancient algebraists.
We make the parallelepiped whose base is the square of ab and height hb, common, so the cube bh plus this parallelepiped equals the (volume of the) cube whose base is ab and height is bc, which we assumed to equal the given number. c h b z d m But the parallelepiped whose base is the square of ab (which is the number of the roots) and whose height is bh, which is the side of the cube, equals the given number of the sides of the cube hb. Hence the cube hb plus the assumed number of its sides equals the given number, which is the desired result.
But the ratio of hm (which equals bt) to mb (which equals ht, which is a line of order of the other section) is equal to the ratio of ht to ab (which is the right side of the conic section). So the four lines are proportional. Hence the ratio of ab to mb is the same as the ratio of mb to bt, and is as the ratio of bt to cm. Now, the ratio of the square of ab (the first) to the square of mb (the second) is the same as the ratio of mb (the second) to cm (the fourth). Hence the cube of mb equals (the volume of) the parallelepiped whose base is the square of ab and height equals cm because their heights are equivalent to their bases.
A Boundary Value Problem in a Strip for Partial Differential Equations in Classes of Tempered Functions by Andryan A. A.