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By Leonard Eugene Dickson

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5. Reduce y 11 = 1 to an equation of degree 5 in x. 44 CONSTRUCTIONS WITH RULER AND COMPASSES [Ch. III 6. Solve y 5 − 7y 4 + y 3 − y 2 + 7y − 1 = 0 by radicals. ] 7. After finding so easily in Chapter I the trigonometric forms of the complex roots of unity, why do we now go to so much additional trouble to find them algebraically? 8. Prove that every real root of x4 + ax2 + b = 0 can be constructed with ruler and compasses, given lines of lengths a and b. 9. Show that the real roots of x3 −px−q = 0 are the abscissas of the intersections of the parabola y = x2 and the circle through the origin with the center ( 12 q, 12 + 21 p).

If, in the general cubic equation x3 + bx2 + cx + d = 0, (1) we set x = y − b/3, we obtain the reduced cubic equation y 3 + py + q = 0, (2) lacking the square of the unknown y , where (3) p=c− b2 , 3 q =d− bc 2b3 + . 3 27 After finding the roots y1 , y2 , y3 of (2), we shall know the roots of (1): (4) b x 1 = y1 − , 3 b x 2 = y2 − , 3 b x 3 = y3 − . 3 43. Algebraic Solution of the Reduced Cubic Equation. We shall employ the method which is essentially the same as that given by Vieta in 1591. We make the substitution (5) y=z− p 3z in (2) and obtain z3 − p3 + q = 0, 27z 3 since the terms in z cancel, and likewise the terms in 1/z .

R6 , we obtain only three distinct values of x: (16) x1 = R+ 1 = R+R6 , R 1 x2 = R2 + 2 = R2 +R5 , R 1 x3 = R3 + 3 = R3 +R4 . ] 41 RECIPROCAL EQUATIONS In order to illustrate a general method of the theory of regular polygons, we start with the preceding sums of the six roots in pairs and find the cubic equation having these sums as its roots. For this purpose we need to calculate x1 + x2 + x3 , x1 x2 + x1 x3 + x2 x3 , x1 x2 x3 . First, by (16), x1 + x2 + x3 = R + R2 + · · · + R6 = −1, since R, .

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