put then a = r cos 0, b = r sin 0, so that r2 = a + b2, tan 0 (a + ib)3 = {r (cos 0 + i sin 0)}3. b Now by De Moivre's theorem the three values of this ex pression are and 1 (cos +ining), (cos ++isin+2), COS 3 where denotes the arithmetical cube root smallest angle found from the equation tan 1 3 The three values of (a - ib)3 are obtained by changing the sign of i in the above results; hence the roots are 581. We shall now give a brief discussion of some of the methods which are employed to obtain the general solution of a biquadratic equation. It will be found that in each of the methods we have first to solve an auxiliary cubic equation; and thus it will be seen that as in the case of the cubic, the general solution is not adapted for writing down the solution of a given numerical equation. 582. The solution of a biquadratic equation was first obtained by Ferrari, a pupil of Cardan, as follows. Denote the equation by x1 + 2px3 + qx2 + 2rx + s 0; add to each side (ax + b)3, the quantities a and b being determined so as to make the left side a perfect square; then x2 + 2px3 + (y + a3) x2 + 2 (r + ab) x + s + b2 = (ax + b)3. Suppose that the left side of the equation is equal to (x2+px+k); then by comparing the coefficients, we have by eliminating a and b from these equations, we obtain or (pk-r)2= (2k+ p2 − q) (k2 — 8), 2k3 − qk2 + 2 (pr − s) k + p3s — qs − 72 = 0. From this cubic equation one real value of k can always be found [Art. 553]; thus a and b are known. (x2 + px + k)2 = (ax + b)2; ... x2 + px + k= = ±(ax+b); Also and the values of x are to be obtained from the two quadratics Add a2x2+2abx+b2 to each side of the equation, and assume x1 − 2x3 + (a2 − 5) x2 + 2 (ab+5) x + b2 − 3 = (x2 − x + k)2; then by equating coefficients, we have By trial, we find that k= -1; hence a2=4, b2=4, ab=−4. Substituting the values of k, a and b, we have the two equations x2-x-1= ±(2x − 2) ; 583. The following solution was given by Descartes in 1637. Suppose that the biquadratic equation is reduced to the form assume x2+qx2 + rx + 8 = 0; x1 + qx2 + rx + 8 = (x2 + kx + 1) (x2 —– kx + m); then by equating coefficients, we have l+m-k2=q, k (m-1)=r, lm=s. From the first two of these equations, we obtain This is a cubic in k2 which always has one real positive solution [Art. 553]; thus when k2 is known the values of and m are determined, and the solution of the biquadratic is obtained by solving the two quadratics x2 + kx+1 = 0, and x2 - kx + m = = 0. Example. Solve the equation Assume x42x2+8x-3=0. x42x2+8x-3= (x2 + kx+1) (x2 – kx+m) ; then by equating coefficients, we have l+m-k2=-2, k (m-1)=8, lm=-3; whence we obtain (k3 − 2k +8) (k3 – 2k – 8) = − 12k2, or k6-4k4+16k2 – 64=0. This equation is clearly satisfied when k2-4=0, or k±2. It will be sufficient to consider one of the values of k; putting k=2, we have Thus hence x1- 2x2+8x-3= (x2+2x − 1)(x2 – 2x+3); and therefore the roots are -1/2, 1-2. 584. The general algebraical solution of equations of a degree higher than the fourth has not been obtained, and Abel's demonstration of the impossibility of such a solution is generally accepted by Mathematicians. If, however, the coefficients of an equation are numerical, the value of any real root may be found to any required degree of accuracy by Horner's Method of approximation, a full account of which will be found in treatises on the Theory of Equations, 585. We shall conclude with the discussion of some miscella neous equations. Example 1. Solve the equations: x+y+z+u=0, ax+by+cz + du=0, a2x + b2y + c2z+d2u=0, a3x+b3y+c3z+d3u = k. Multiply these equations, beginning from the lowest, by 1, p, q, r respectively; p, q, r being quantities which are at present undetermined. Assume that they are such that the coefficients of y, z, u vanish; then Thus the value x is found, and the values of y, z, u can be written down by symmetry. Thus the value of x is found, and the values of y, z, u can be written down by symmetry. The solution of the above equations has been facilitated by the use of Undetermined Multipliers. Example 2. Shew that the roots of the equation (x-a) (x-b) (x-c) -f2(x-a)-g2(x-b) - h2 (x-c)+2fgh=0 are all real. From the given equation, we have (x-a){(x-b) (x-c)-ƒa} - {g3 (x - b) + h2 (x - c)-2fgh}=0. Let p, q be the roots of the quadratic (x-b)(x-c)-f2=0, and suppose p to be not less than q. By solving the quadratic, we have 2x=b+c±√(b − c)2 +4ƒ2. ...(1); now the value of the surd is greater than bc, so that p is greater than b and q is less than b or c. or c, In the given equation substitute for a successively the values the results are respectively since -(g√p-b-hp-c)2, +(g√b-q-h√c-g)2, -∞, Thus the given equation has three real roots, one greater than p, one between Ρ and q, and one less than q. If p-q, then from (1) we have (b − c)2+4ƒ2=0 and therefore b=c, f=0. In this case the given equation becomes (x − b) { (x − a) (x − b) − g2 – h2} = 0; thus the roots are all real. If p is a root of the given equation, the above investigation fails; for it only shews that there is one root between q and +, namely p. But as before, there is a second real root less than q; hence the third root must also be real. Similarly if q is a root of the given equation we can shew that all the roots are real. The equation here discussed is of considerable importance; it occurs frequently in Solid Geometry, and is there known as the Discriminating Cubic. 586. The following system of equations occurs in many branches of Applied Mathematics. y + + 2 1. (0-λ) (0-μ) (0 – v). a+o b+0 c+6 (a+0) (b+0) (c+0)' x, y, z being for the present regarded as known quantities, |