We have that is x (x2+y3+z3-3xyz)=a2-bc. Hence, from the last equation and the two similar ones, It is obvious that x=a, y=b, z=c will satisfy the equations: put then x=a+, y=b+μ, z=c+v, and we have after reduction = x=a, y=b, z=c; 2 (b−c) (ca) (a – b) = = a (b−c) ̄ ̄b (c- a) b(c-a) c (a - b) ̄ ̄ a2 (b−c)2+b2 (c − a)2 + c2 (a - b)2* Ex. 7. Solve the equations x+y+z = 6, yz+zx+xy=11, xyz= 6. This is an example of a system of three symmetrical equations. Such equations can generally be easily solved by making use of the relations of Art. 129. Thus in the present instance it is clear that x, y, z are the three roots of the cubic equation The roots of the cubic are 1, 2, 3. Hence x=1, y=2, z=3; or x=1, y=3, z=2; y=3, z=1; &c. or x=2, ..(i), ..(ii), .(iii). This again is a system of symmetrical equations, and two of the relations of Art. 129 are already given; we have therefore only to find the third. Then, from (i), (iii) and (iv), we see that x, y, z are the roots of the cubic that is ... λ3 - aλ2 - c2λ + ac2=0, λ2 (λ − a) - c2 (λ − a) = 0 ; λ=a, or λ= ±c. Thus xa, y=c, z=−c; &c. x2 (y-2)+ y2 (z-x) + z2 (x − y) = a2 (b−c) +b2 (c− a)+c2 (a - b), that is (y-2) (2−x) (x − y) = (b−c) (ca) (ab). By multiplication x2y2z2 (y − z) (z − x) (x − y) = a2b2c2 (b − c) (c − a) (a − b); Again a2 (b-c) y + b2 (c − a) x = x2y (y − z)+xy2 (z — x) ..(i), .(ii). .(iii). |