29. The multiplication of one expression by another may be effected by a series of distributions, and when the expressions are not too complex this is usually the best method.

Thus (a+b) (a− b + c) = a(a − b + c) + b (a−b+c) = a2 — ab + ac + ab − b2 + bc = a2 − b2 + ac + be.

By remembering and applying a few elementary product forms, the operation may often be much curtailed. As convenient fundamental forms we may take the following, although any simple form that can be remembered may be equally useful:

(1) (a + b)2 = a2 + b2 + 2 ab. (2) (a - b)2 = a2 + b2 — 2 ab.

(3) (a−b) (a + b) = a2 — b2.

(4) 3 (a + b) (b + c) (c + a) = (a + b + c) 3 — a3 — b3 — c3 = (Σα) - Σαβ.

(5) (a+b+c) (a2 + b2 + c2 — ab- beca)

or

= a+b+c3-3 abc,

Σα (Σα? - Σαb) = Σα - 3 abc.

The mark .. is a verbal symbol for 'therefore' or 'hence.'

Ex. 1. (a + b c + d) (a + b + c - d)

=(a+bcd) (a + b + c − d)

= (a + b)2 — (c — d)2

= a2 + b2 — c2 – d2 + 2 ab + 2 cd.

Ex. 2. (a + b + c + d)2 = (a + b)2+ 2(a + b) (c + d) + (c + d)2 = a2 + b2 + c2 + d2 + 2(ab + ac + ad + bc + bd + cd).

Ex. 3. To distribute

s(-a) (sb)+s(s—b) (s−c)+s(sc) (sa) - (s—a) (s—b) (s—c), where 2s = a + b + c. s, being symmetrical in a, b, and c, is not altered by a circular substitution of these letters.

But this substitution brings s(s - a) (s — b) to s(s — b) (s — c), and s(sb) (s — c) to s(s – c) (s − a).

Hence having the expansion of s(s − a)(s — b), the expansions of the two following terms may be immediately written down by a cyclic interchange of letters.

3 s3 − 2 s2(a + b + c) + s(ab + bc + ca),

and as 2s = a+b+c, this becomes

or

Again the expansion of (s — a) (s — b) (s — c) is

s3 − s2(a + b + c) + s(ab + bc + ca) — abc,

· 83 + s(ab + bc + ca) — abc.

And subtracting this from A gives abc as the final result.

А

This example furnishes a good illustration of the remark in Art. 20, for the long series of operations symbolized in the statement of the exercise is just equivalent in its totality to the two multiplications symbolized in the final result.

Ex. 4. To prove the identity

8(Σa)3 — Σ(a + b)3 = 3(2 a + b + c) (2 b + c + a) (2 c + a + b). Either of two methods may be adopted—(1) to transform one member to the other by the rules of operation; or (2) to transform each to the same third expression by distribution. We shall adopt the first way.

8(a)=(2a)3 = (a + b + b + c + c + a)3.

Now put a + b = p, b + c = q, c + a = r, and the identity reduces to

(p + q + r)3 − p3 — q3 — r3 = 3(p + r) (r + q) (I +p), which is true by Art. 29, (4).

30. Expansion of Symmetrical Homogeneous Expressions.

This form of expression is of frequent occurrence, and its properties of symmetricality and homogeneity enable us to expand it with some facility.

Ex. 1. To expand (a + b + c)3 − (a + b)3 − (b + c)3 − (c + a)3. Being homogeneous and of 3 dimensions, the type terms in its expansion can only be a3, a2b, and abc. Taking the type a3, we see that its coefficient is - 1. Taking the type a2b, its coefficient is readily found to be zero. And the coefficient of the type abc is 6.

Ex. 2. To expand (Σa)3 + (a + b − c) (b + c − a) (c + a − b). The expansion being homogeneous of 3 dimensions and symmetrical, must be of the form

EXERCISE II. b.

1. Write out the type terms in the following symmetrical and homogeneous expressions

(a + b−c)(b + c − a) (c + a−b)

= ab(a+b)+bc(b + c) + ca(c + a) — a3 — b3 — c3 – 2 abc.

3. Za Zab-(a + b) (b + c) (c + a) =abc.

4. Show that {(a − b) (2 b − c)} = (Za)2 — 3 Za2.

5. Show that

(a+b+c) (a + b −c) (b + c − a) (c + a − b)

=2Za2b2 - Za1= a2 (2 b2 — a2) + b2 (2 c2 — b2) + c2 (2 a2 — c2)

iii. Za3(b −c) – Za · Za2(b − c).

iv. a(sb)(s— c) + b ( s − c) ( s − a) + c( s − a) (s — b)

+ 2(s − a) (s — b)(s — c), where 2 s = a + b + c.

7. If 2s = a+b+c, show that the three following expressions are identical in value

s(sa) (b+c) + a (s - b) (s - c) - 2 bcs,

8. Show that (x − b) (x − c) (b − c) + (x − c) (x − a) (c − a) + (x − α) (x − b) (a − b) + (a − b) (b − c) (c − a) =0.

9. Show that (2a)2 = Za2 + 2 Zab, with any number of letters. 10. Multiply Za2 + 2 Zab by Za, 3 letters.

11. Multiply Za2 + Zab by Zab, 3 letters.

12. Multiply Za2+2 Zab by Za, 4 letters. This gives (a + b + c + d)3.

Prove the following theorems in numbers

13. The difference between the square of the sum and the square of the difference of two numbers is the product of twice the numbers.

14. The sum of the squares of the sum and of the difference of two numbers is one-half the sum of the squares of twice the numbers.

15. If two numbers be each the sum of two squares, their product is the sum of two squares.

16. If two numbers be each the difference between two squares, their product is the difference between two squares.

17. If the sum of two numbers is 1, their product is equal to the difference between the sum of their squares and the sum of their cubes.

18. If the product of two numbers is 1, the square of their sum exceeds the sum of their squares by 2.

31. The distribution of (x+a) (x + b) (x + c), the product of a number of binomial factors with one letter the same in each factor, is very important.

The dominant letter, x, is taken as the variable, and the expansion is arranged according to the powers of x. With three factors we readily see that taking the from every factor gives 3; taking it from every factor