A method for removing all intermediate terms from a given equation

Abstract

We have learned from DesCartes’ geometry by what method the second term might reliably be removed from a given equation; but on the question of removing multiple intermediate terms I have seen nothing hitherto in the analytic arts. On the contrary, I have encountered not a few who believed that the thing could not be done by any art. For this reason I have decided to set down here some things concerning this business, enough at least for those who have some grounding in the analytic art, since the others could scarcely be content with so brief an exposition: reserving the remainder (which they might wish to see here) for some other time. Thus, in the first place, this must be noted; let some given cubic equation be x3− px2 +qx− r = 0, in whichx signifies the roots of this equation; and p, q, r represent known quantities. Now in order to remove the second term, let us suppose that x = y+a; now with the aid of these 2 equations, a third may be discovered in which the quantity x is absent, and this will be 1 y3 z3ayy z3aay za3 = 0 −pyy −2pay −paa zqy zqa −r Now let the second term be made equal to zero (since this is the term we intend to remove) and we shall have 3 ay2− py2 = 0. Whencea= p 3 : which shows that, for removing the second term in the cubic equation, x = y+ a is to be replaced (as we have just done) by y = x+ p 3 . These things have been fully published, nor are they here spoken of for any other reason than that they serve to illustrate what is to follow, since when these things have been fully understood it is easier to grasp those things I am now about to propose. Secondly, now let there be two terms to be removed from a given equation, and I say that we must suppose x2 = bx+y+a; if three,x3 = cx2 +bx+y+a, if four x4 = dx3 +cx2 +bx+y+a and so to infinity. But I shall call theseassumed equations, in order to distinguish them from equations which may be considered as given. The reason for this is that, for the same reason that with the help of the equation x = y+a only one term at least can be removed (because of course at least only one indeterminant a exists), so by the same reason by the help of x2 = bx+y+a only two terms can be removed because just two indeterminants a andb are present; and so furthermore with the help of the followingx3 = cx2 +bx+y+a not more than three can be removed since there are only three indeterminants a, b, c. But since it may be understood by what reason this should follow, I shall show by what reason two terms may be removed from a given equation by means of the assumed [equation] x2 = bx+y+a, and from this it will be easily established by what method one must proceed in this matter however far one might wish (since one may everywhere proceed by the same method). So be it. Thirdly, [there is] the cubic equation x3− px2 + qx− r = 0, from which the two intermediate terms are to be removed: first, the second term is removed (which is certainly of no help, but may at least be omitted here for the sake of time), and then we shall obtain an equation like this y3− qy− r = 0. Now let the assumed equation (in accordance with our second paragraph) be y2 = by+z+a and following from this let there be a third equation (by proceeding according to the recognized rules of the analysts) in which the quantity y is absent, and we shall obtain