And consequently, by taking the latter of these series from the former, and making the first term of the result a multiplier, we shall have &c.}. But since s= √ (‡b2— ‚1, a3), we shall have 6 + 2762 2762-4a3 Whence, if these values be substituted for their equals in the last series, there will arise the above expression for the first root of the equation. And if we put the root thus found, or its equivalent expression 3/ { √(162— 21, a3) +16 } — 3/{√ (462— 1, a3) — 16 } =±r, 27 b we shall have, according to the formula before given for the other two roots, 6]}. Or, taking, as before, √ (1b2—1, b3)=s, and simplifying +2 V 27 the result, a === (1 + 1 ) + & (1 − 9 ) } . 2s Whence, by extracting the roots of the righthand side of this equation, there will arise &c. 2 &c. And, consequently, if the latter of these series be added to the former, we shall have, by making the first term of the result a multiplier, D-, &c., which series also answers to the irreducible case, and must be used when 2a3 is greater than 2762. And if the root thus found, be put =r, as before, the other two 4a3-27b2 4 3.6.9.12 4a3 — 2762)2 + -, &c.}. Or, 4a3 2762 {1+ 2.5.8.11.14 3.6.9.12.15.18 4 { 2 2762 5.8 1 + A 3.64a3 2762 9.12 11.14 2762 17.20 2762 21.24 4a3 — 27b2) &c. Where the signs are to be taken as in the latter part of the preceding case. EXAMPLES. 1 Given 3+6x= 2, to find the value of x. Here a= 6, and b = 2, whence Hence, if these values be substituted for their equals in the above series, the result will give the above expressions for the two remaining roots of the equation. |