Steps / Historical AI

Leonhard Euler · 1770 (1828 English ed.) · Internet Archive

Elements of Algebra

All 627 leaves · leaf 119

Transcription

82 ELEMENTS SECT. II. $a^2+2ab+2b^2$ $a^2-2ab+2b^2$ --- $a^4+2a^3b+2a^2b^2$ $-2a^3b-4a^2b^2-4ab^3$ $2a^2b^2+4ab^3+4b^4$ --- $a^4+b^4$ --- $2a^2-3ab-4b^2$ $3a^2-2ab+b^2$ --- $6a^4-9a^3b-12a^2b^2$ $-4a^3b+6a^2b^2+8ab^3$ $2a^2b^2-3ab^3-4b^4$ --- $6a^4-13a^3b-4a^2b^2+5ab^3-4b^4$ --- $a^2+b^2+c^2-ab-ac-bc$ $a+b+c$ --- $a^3+ab^2+ac^2-a^2b-a^2c-abc$ $a^2b+b^3+bc^2-ab^2-abc-b^2c$ $a^2c+b^2c+c^3-abc-ac^2-bc^2$ --- $a^3-3abc+b^3+c^3$ --- 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, viz. I. $(a+b)$ II. $(a^2+ab+b^2)$ III. $(a-b)$ IV. $(a^2-ab+b^2)$. 1st. The product of the factors I. and II. $a^2+ab+b^2$ $a+b$ --- $a^3+a^2b+ab^2$ $+a^2b+ab^2+b^3$ --- $a^3+2a^2b+2ab^2+b^3$ 2d. The product of the factors III. and IV. $a^2-ab+b^2$ $a-b$ --- $a^3-a^2b+ab^2$ $-a^2b+ab^2-b^3$ --- $a^3-2a^2b+2ab^2-b^3$
confidence 1.00 32 expressions source scan at the Internet Archive

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