This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1904 Excerpt: ...P = (a + b + c-)8, we have P = a3 + (3a1 + 3ab + b)b + 3(a + by + 3(a + b)c + c22c+--; where the leading terms of the several groups on the right, namely, a?, 3 ab, 3 a-, are all of higher degree in x than any of the terms which follow them. From this identity we may find a, b, c, as follows: 1. Evidently a is the cube root of the leading term of P. 2. Subtract a8 from P. As the leading term of the remainder, Ru must equal 3a%, we may find b by dividing this term by 3 a2. 3. Having found b, form (3 a2 + 3 ab + J2) b and subtract it from Rv As the leading term of the remainder, Rt) must equal 3 a?c, we may find c by dividing this term by 3 a2. 4. Continue thus until a remainder is reached which is of lower degree than a2. If this final remainder is 0, then P is, as was supposed, a perfect cube and its cube root is a + b + c +. If this final remainder is not 0, P is not a perfect cube, but we shall have reduced it to the form P = (a + b + c +---y + R, where R is of lower degree than a2. It is convenient to arrange this reckoning as follows: Example. Find the cube root of a» + 6x5 + 21a + 44x3 + 63x2 + 54x + 27. x2+ 2x + 3 3a2 = 3x4 x6 + 6x5 + 21x4 + 44x3+ 63x2 + 54x + 27 x6 3(x2)2 = 3x Since the final remainder is 0, x6 + 6 Xs + + 54 x + 27 is a perfect cube and its cube root is x2 + 2 x + 3. Compare § 569, Ex. 2. Observe that as each new remainder Ru R1 is found, we divide its leading term by 3 a'2 and so get the next term of the root. Then at the left of the remainder we write the sum of three times the square of the part of the root previously obtained, three times the product of this part by the new term, and the square of the new term. We multiply this sum by the new term, subtract the result from the remainder under consideration, and thus obt...
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