
308 H. F. TALBOT ON A GENERAL SOLUTION OF 
Determine P from the formula p = 38n? — 2an + 6, making P the reduct of p. 
Substitute » for the unknown quantity x in the original equation (not in the 
reduct equation), and let the result be 9% I put the result in this form, be- 
cause it will always be divisible by 9 (since by hypothesis x satisfies the reduct 
equation). Therefore divide by 9 and we get &. And now let the reduct of k be 
called Q. 
Solve the equation Py = Q, where P and Q are of course numbers not ex- 
ceeding nine, and thus y becomes known. Finally, put 2 = 9y + » and «# will 
be one of the roots of the given cubic. 
This will best be illustrated by examples. 
First Example. 
Let the given equation be a’— 162 2? + 8531 « — 146370 =0. Its reduct 
will be 2 — 02? + 8 —3 = 0, ora’® + 8a—3 =0, so that a=0, b= 8. 
Substitute for a the numbers one to eight, and by help of Table A we get 
8x 
81 
82° 
8°3 
84 
85 
8'6 
8°7 
8'8 — 3 e 
This shows that the reduct equation can only be satisfied by taking either 5, 6, 
or 7 as the value of 7, and in fact we shall find that the three roots of the cubic 
are of the forms 9y +5, 9y'+6, 9y' +7. To test the accuracy of this, 
multiply together (w — 5) (wv — 6) (vw — 7) = 0, and we get the reduct equation 
x® + 8x —3 = 0, as before. 
Let us now calculate the three roots of the cubic in succession. 
First, take the root 2 = 9y + 5 so that m = 5, then the formula p = 3n? — 
2an + b becomes p = 3n’ + 8 (sincea=0, b= 8). Hence p=3.25+8=83, 
whose reduct is 2. Therefore P= 2. Next, substitute 5 for the unknown 
quantity in the original equation a — 162 a + 8531 w — 146370, and the 
result is nine times 11960, the reduct of which number is 8: we have therefore 
Q=8. We now take the equation Py = Q or 2y = 8, which gives us y = 4. 
Finally, we have « = 9y + n= 9y +5=9.4+5=41. Therefore 41 is one 
of the roots of the equation a —162 a? + 8531 # — 146370 =0. It is 
evident that it would be difficult to obtain this root by mere guessing, and it 
cannot be fuund by Carpan’s rule, therefore a direet mode of solution is to be 
welcomed. 
_ Let us now proceed to find the second root. Let m = 6, then the formula 
p = 3n? + 8 becomes p 116 whose reduct is 8, therefore P = 8. Substitute 
With number 1 — 3 for the reduct. 
% 3 
3 3 
2? 4 3 ” 
5 0 
6 0 
7 0 
+++++4+4+4]+ 
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