93. Symmetrical expressions. An expression which is unaltered by interchanging any pair of the letters which it contains is said to be a symmetrical expression. Thus a+b+c, bc + ca + ab, a3 + b3 + c3-3abc are symmetrical expressions. It is clear that the product of two symmetrical expressions is symmetrical, for if neither of two factors is altered by an interchange of two letters their product cannot be altered by such interchange. It is also clear that the quotient of two symmetrical expressions is a symmetrical expression. Expressions which are unaltered by a cyclical change of the letters involved in them are called symmetrical expressions, although they may not satisfy the condition. of being unaltered by an interchange of any two of the letters. For example, the expression (b − c) (c − a) (a − b) is considered to be a symmetrical expression since it is unaltered by changing a into b, b into c and c into a; but by interchanging two of the letters a result is obtained which differs in sign from the original expression. Ex. 1. Find the factors of a2 (b − c) + b2 (c − a) +c2 (a − b). If we put bc in the expression a2 (b-c)+b2 (c-a)+c2 (a - b)....... it is easy to see that the result is zero. ..(i) Hence bc is a factor of (i), and we can prove in a similar manner that c-a and a-b are factors. Now (i) is an expression of the third degree; it can therefore only have three factors. Hence (i) is equal to 1 2. L (b−c) (c - a) (a - b)............. .(ii), where L is some number, which is always the same for all values of a, b, c. By comparing the coefficients [See Art. 91] of a2 in (i) and (ii) we see that L= -1. We can also find L by giving particular values to a, b and c. Thus, let a=0, b=1, c=2; then (i) is equal to -2, and (ii) is equal to 2L, and hence as before L= -1. Ex. 2. Find the factors of a3 (b − c) + b3 (c − a) + c3 (a - b). As in the preceding example, (b − c), (c − a) and (a - b) are all factors of a3 (b-c)+b3 (c-a)+c3 (a - b)......................(i). Now the given expression is of the fourth degree; hence, besides the three factors already found, there must be one other factor of the first degree, and this factor must be symmetrical in a, b, c, it must therefore be a+b+c. Hence the given expression must be equal to L(b-c) (c-a) (a - b) (a+b+c)...... where L is a number. .(ii), By comparing the coefficients of a3 in (i) and in (ii) we see that L= -1; hence a3 (b−c) + b3 (c− a) + c3 (a - b) = − (b − c) (c − a) (a - b) (a+b+c). We can also find L by giving particular values to a, b, and c. Thus, let a=0, b=1, c=2; then (i) is equal to 6 and (ii) is equal to 6L, so that L= -1. We may also proceed as follows: Arrange the expression according to powers of a; thus a3 (b −c) - a (b3 − c3) + bc (b2 — c2). It is now obvious that b-c is a factor, and we have 1 − (b − c) (c − a) {b2 + bc − a2 — ac } = − (b − c) (c − a) (a − b) (a+b+c). = Ex. 3. Find the factors of b2c2 (b − c) + c2a2 (c − a) + a2b2 (a − b). b2c2 (b−c) + c2a2 (c − a) + a2b2 (a - b). ..............(i), it is easy to see that the result is zero; hence b-c is a factor of (i). So also ca and a- b are factors. The given expression being of the fifth degree, there must be, besides the three factors bc, c- a, ab, another factor of the second degree; also, since this factor must be symmetrical in a, b, c, it must be of the form L (a2+b2+c2) + M (bc+ca+ab). Thus (i) is equal to (b−c) (ca) (a - b) {La2 + Lb2 + Lc2+ Mbc + Mca+Mab}...(ii). Equating coefficients of a1 in (i) and in (ii) we see that L = 0; and then equating coefficients of b3c2 we see that M = -1. Hence (i) is equal to - (b-c) (ca) (a - b) (bc+ca + ab). We may also proceed as follows. Knowing that b-c, c-a, and a -b are factors, it is easy to group the terms so as to shew the successive quotients. 10. (a+b+c)3 − (b + c − a)3 − (c + a − b )3 − (a + b − c)3. 11. (a + b + c)3 − (b + c − a)3 − (c + a − b) 5 − (a + b −c)3. 12. a (b+c-a)2 + b (c + a − b )2 + c (a+b−c) + (b + c − a) (c + a−b) (a + b −c). 13. a3 (b+c-a) + b2 (c + a − b) + c2 (a + b − c) -(b+c-a) (c+a−b) (a + b−c). 14. (b+c− a) (c + a − b ) (a + b − c) + a ( a − b + c) (a + b − c) +b(a+b-c) (a+b+c) + c (a+b+c) (a−b+c). 15. (b-c) (a−b + c) (a + b −c) + (c − a) (a + b − c) (− a+b+c) + (a−b) (− a + b + c) (a − b + c). 16. (x + y + z)3 — x3 — y3 — z3. 17. (x + y + z)3 — x3 — y3 — z3. (b-c) (b+c)2 + (c − a) (c + a)2 + (a − b) (a + b)2. - 18. 19. (b−c) (b+c)3 + (c − a) (c + a)3 + (a − b) (a + b)3. 20. (b−c) (b+c)2 + (c − a) (c + a)2 + (a − b) (a + b)1. a3 + b3 + c3 +5abc-a (a - b) (a–c) - b (b-c) (b-a) - c(c− a) (c-b). 21. 22. a2 (a+b) (a + c) (b − c) + b2 (b + c) (b + a) (c − a) 23. (y + z)(x+x) (x + y) + xyz. +c2 (c+a) (c+b) (a – b). 24. a3 (b+c)3 + b2 (c + a)2 + c2 (a + b)2 + abc (a + b + c) +(a2 + b2 + c2) (bc + ca + ab). 25. (x + y + z)* - (y + x)1 − (≈ + x)1 − (x + y)* + x2 + y2+za. 26. a2 (b+c-2a) + b2 (c + a-2b) + c2 (a + b - 2c) + 2 (c2 − a3) (c − b) + 2 (a2 − b3) ( a − c) + 2 (b2 — c2) (b − a). 27. (b+c-a-d)1 (b − c) (a − d) + (c + a − b − d)* (c − a) (b −d) +(a+b-e-d)* (a - b) (c–d). 28. Shew that 12 {(x + y + z)2" − (y + z)2n − (≈ + x)21 − (x + y)11 + x2 + y2n + z2n} is divisible by (x + y + z)* − (y + z)* − (≈ + x)* − (x + y)* + xa +y* + z*. 29. Shew that 1 2 a3 (b + c − a)2 + b3 (c + a − b)2 + c3 (a + b −c)2 + abc (a2 + b2 + c2) + (a2 + b2 + c2 − bc — ca − ab) (b + c − a) (c + a − b ) (a + b − c) 30. Shew that (b − c)® + (c − a)® + (a − b)o − 9 (b − c)3 (c − a)3 (a − b)3 = 2 (a − b)3 (a − c)3 + 2 (b − c)3 (b − a)3 + 2 (c − a)3 (c − b)3. 31. Shew that (b + c)3 + (c + a)3 + (a + b)3 + (a + d)3 + (b + d)3 + (c + d)3 32. 33. =3(a+b+c+d) (a2 + b2 + c2 + d3). Reduce to its simplest form 4 (a2 + ab + b2)3 − (a − b)3 (a + 2b)3 (2a + b)3. Shew that a* (b2 + c2 − a2)3 + b1 (c2 + a2 − b2)3 + c* (a2 + b2 − c2)3 is divisible by 34. Resolve into quadratic factors 4{cd (a2 — b3) + ab (c2 — d2)}2 + {(a2 — b3) (c2 – d3) — 4abcd}3. 35. Shew that (y2 -- x2) (1 + xy) (1 + xz) + (≈2 − x2) (1 + yz) (1 + yx) + (x2 - y3) (1 + zx) (1 + zy) = (y − ≈) (≈ − x) (x − y) (xyz + x + y + z). 36. Find the factors of a3 (b − c) (c – d) (d — b) — b3 (c — d) (d − a) (a − c) + 37. c3 (d − a) (a − b) (b − d) – d3 (a – b) (b − c) (c − a). Find the factors of b3c3d3 ( b − c) (c — d) (d — b) — c2d3a2 (c — d) (d − a) (a − c) + d3a3b2 (d − a) (a - b) (b − d) — a2b3c2 (a − b) (b − c) (c − a). |