### Part I. Section II. Chapter 3. “Of the Multiplication of Simple Quantities.”

270 When it is only required to represent multiplication, we put each of the expressions, that are to be multiplied together, within two parentheses, and join them to each other, sometimes without any sign, and sometimes placing a point ⋅ between them.

Thus, for example, to represent the product of the two expressions $$a-b+c$$ and $$d-e+f$$ we write

$(a-b+c) \cdot (d-e+f)$

or barely,

$(a - b + c) (d-e+f)$

which method of expressing products is much used, because it immediately exhibits the factors of which they are composed.

271 But in order to show how multiplication is actually performed, we may remark, in the first place, that to multiply, for example, a quantity, such as $$a-b+c$$, by 2, each term of it is separately multiplied by that number; so that the product is

$2a-2b+2c.$

And the same thing takes place with regard to all other numbers; for if $$d$$ were the number by which it was required to multiply the same expression, we should obtain

$ad-bd+cd.$

272 In the last article, we have supposed $$d$$ to be a positive number; but if the multiplier were a negative number, as $$-e$$, the rule formerly given must be applied; namely, that unlike signs multiplied together produce -, and like signs +. Thus we should have

$-ae+be-ce.$

273 Now, in order to show how a quantity, $$A$$, is to be multiplied by a compound quantity, $$d-e$$; let us first consider an example in numbers, supposing that $$A$$ is to be multiplied by 7-3. Here it is evident, that we are required to take the quadruple of $$A$$: for if we first take $$A$$ seven times, it will then be necessary to subtract $$3A$$ from that product.

In general, therefore, if it be required to multiply $$A$$ by $$d-e$$, we multiply the quantity $$A$$ first by $$d$$, and then by $$e$$, and subtract this last product from the first: whence results $$dA-eA$$.

If we now suppose that $$A=a-b$$, and that this is the quantity to be multiplied by $$d-e$$; we shall have

$\begin{gather} dA=ad-bd\\ eA=ae-be \end{gather}$

whence $$dA-cA=ad-bd-ae+be$$ is the product required.

274 Since therefore we know accurately the product $$(a-b)\cdot (d-e)$$, we shall now exhibit the same example of multiplication under the following form:

$\begin{array}{rrrr} a&-b&&\\ d&-e&&\\ \hline ad&-bd&-ae&+be. \end{array}$

Which shows, that we must multiply each term of the upper expression by each term of the lower, and that, with regard to the signs, we must strictly observe the rule before given; a rule which this circumstance would completely confirm, if it admitted of the least doubt.

275 It will be easy, therefore, according to this method, to calculate the following example, which is, to multiply $$a+b$$ by $$a-b$$;

$\begin{array}{rrr} a&+b&\\ a&-b&\\ \hline a^2&+ab&\\ &-ab&-b^2\\ \hline \textrm{Product}&a^2&-b^2. \end{array}$

276 Now, we may substitute for $$a$$ and $$b$$ any numbers whatever; so that the above example will furnish the following theorem; namely, The sum of two numbers, multiplied by their difference, is equal to the difference of the squares of those numbers: which theorem may be expressed thus:

$(a+b)(a-b)=a^2-b^2.$

And from this another theorem may be derived; namely, The difference of two square numbers is always a product, and divisible both by the sum and by the difference of the roots of those two squares; consequently, the difference of two squares can never be a prime number.

277 Let us now calculate some other examples:

I)

$\begin{array}{rrr} 2a&-3&\\ a&+2&\\ \hline 2a^2&-3a&\\ &4a&-6\\ \hline 2a^2&+a&+6 \end{array}$

II)

$\begin{array}{rrrr} 4a^2&-6a&+9&\\ 2a&+3&&\\ \hline 8a^3&-12a^2&+18a&\\ &+12a^2&-18a&+27\\ \hline 8a^3&+27&& \end{array}$

III)

$\begin{array}{rrrr} 3a^2&-2ab&-b^2&\\ 2a&-4b&&\\ \hline 6a^3&-4a^2b&-2ab^2&\\ &-12a^2b&+8ab^2&+4b^3\\ \hline 6a^3&-16a^2b&+6ab^2&+4b^3 \end{array}$

IV)

$\begin{array}{rrrrr} a^2&+2ab&+2b^2&&\\ a^2&-2ab&+2b^2&&\\ \hline a^4&+2a^3b&+2a^2b^2&&\\ &-2a^3b&-4a^2b^2&-4ab^3&\\ &&+2a^2b^2&+4ab^3&+4b^4\\ \hline a^4&+4b^4&& \end{array}$

V)

$\begin{array}{rrrrr} 2a^2&-3ab&-4b^2&&\\ 3a^2&-2ab&+b^2&&\\ \hline 6a^4&-9a^3b&-12a^2b^2&&\\ &-4a^3b&+6a^2b^2&+8ab^3&\\ &&+2a^2b^2&-3ab^3&-4b^4\\ \hline 6a^4&-13a^3b&-4a^2b^2&+5ab^3&-4b^4\\ \end{array}$

VI)

$\begin{array}{rrrrrrrrrr} a^2&+b^2&+c^2&-ab&-ac&-bc&&&&\\ a&+b&+c&&&&&&&\\ \hline a^3&+ab^2&+ac^2&-a^2b&-abc&-a^2c&-abc&&&\\ &-ab^2&&+a^2b&&-abc&+b^3&+bc^2&-b^2c&\\ &&-ac^2&&+a^2c&-abc&&-bc^2&+b^2c&+c^3\\ \hline a^3&-3abc&+b^3&+c^3&&&&&& \end{array}$

278 When we have more than two quantities to multiply together, it will easily be understood that, after having multiplied two of them together, we must then multiply that product by one of those which remain, and so on: but it is indifferent what order is observed in those multiplications.

Let it be proposed, for example, to find the value, or product, of the four following factors, namely

I. II. III. IV.
$$(a+b)$$ $$(a^2+ab+b^2)$$ $$(c-d)$$ $$(a^2-ab+b^2)$$

First, multiply factors I. and II.:

$\begin{array}{lrrrr} \textrm{II.}&a^2&+ab&+b^2&\\ \textrm{I.}&a&+b&&\\ \hline &a^3&+a^2&+ab^2&\\ &+a^2b&+ab^2&+b^3\\ \hline \textrm{I. II.}&a^3&+2a^2b&+2ab^2&+b^3 \end{array}$

Then multiply factors III. and IV.:

$\begin{array}{lrrrr} \textrm{IV.}&a^2&-ab&+b^2&\\ \textrm{III.}&a&-b&&\\ \hline &a^3&-a^2&+ab^2&\\ &-a^2b&+ab^2&-b^3\\ \hline \textrm{III. IV.}&a^3&-2a^2b&+2ab^2&-b^3 \end{array}$

It remains now to multiply the first product I. II. by this second product III. IV.

$\begin{array}{rrrrrrr} a^3&+2a^2b&+2ab^2&+b^3&&&\\ a^3&-2a^2b&+2ab^2&-b^3&&&\\ \hline a^6&+2a^5b&+2a^4b^2&+a^3b^3&&&\\ &-2a^5b&-4a^4b^2&-4a^3b^3&-2a^2b^4&&\\ &&2a^4b^2&+4a^3b^3&+4a^2b^4&2ab^5&\\ &&&-a^3b^3&-2a^2b^4&-2ab^5&-b^6\\ \hline a^6&-b^6&&&&& \end{array}$

which is the product required.

279 Now let us resume the same example, but change the order of it, first multiplying the factors I. and III. and then II. and IV. together:

$\begin{array}{rrrrrrrr} \textrm{I. II.}&a^3&+2a^2b&+2ab^2&+b^3&&&\\ \textrm{III. IV.}&a^3&-2a^2b&+2ab^2&-b^3&&&\\ \hline &a^6&+2a^5b&+2a^4b^2&+a^3b^3&&&\\ &-2a^5b&-4a^4b^2&-4a^3b^3&-2a^2b^4&&\\ &&+2a^4b^2&+4a^3b^3&+4a^2b^4&+2ab^5&\\ &&&-a^3b^3&-2a^2b^4&-2ab^5&-b^6\\ \hline &a^6&-b^6&&&& \end{array}$

which is the product required.

280 We may perform this calculation in a manner still more concise, by first multiplying the factor I. by factor IV. and then factor II. by factor III.

$\begin{array}{rrrr} \textrm{I.}&a&+b&\\ \textrm{III.}&a&-b&\\ \hline &a^2&+ab&\\ &&-ab&-ab^2\\ \hline \textrm{I. III.}&a^2&-b^2& \end{array}$
$\begin{array}{rrrrrr} \textrm{II.}&a^2&+ab&b^2&&\\ \textrm{IV.}&a^2&-ab&+b^2&&\\ \hline &a^4&+a^3b&+a^2b^2&&\\ &&-a^3b&-a^2b^2&-ab^3&\\ &&&+a^2b^2&+ab^3&+b^4\\ \hline \textrm{II. IV.}&a^4&+a^2b^2&+b^4&&\\ \end{array}$

It remains to multiply the product I. IV. by that of II. and III.

$\begin{array}{rrrr} \textrm{I. IV.}&a^3&+b^3&\\ \textrm{II. III.}&a^3&-b^3&\\ \hline &a^6&+a^3b^3&\\ &&-a^3b^3&-b^6\\ \hline &a^6&-b^6& \end{array}$

the same result as before.

281 It will be proper to illustrate this example by a numerical application. For this purpose, let us make $$a=3$$ and $$b=2$$, we shall then have $$a+b=5$$, and $$a-b=1$$; farther, $$a^2=9$$, $$ab=6$$, and $$b^2=4$$: therefore $$a^2+ab+b^2=19$$, and $$a^2-ab+b^2=7$$: so that the product required is that of

5 ⋅ 19 ⋅ 1 ⋅ 7, which is 665.

Now, $$a^6=729$$, and $$b^6=64$$; consequently, the product required is $$a^6-b^6=665$$, as we have already seen.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.