Elements of Algebra

108 ELEMENTS SECT. II. We see therefore that it contains the cubes of the two parts of the root, and, beside that, $3a^2b+3ab^2$; which quantity is equal to $(3ab) \times (a+b)$; that is, the triple product of the two parts, $a$ and $b$, multiplied by their sum. 334. So that whenever a root is composed of two terms, it is easy to find its cube by this rule: for example, the number $5=3+2$; its cube is therefore $27+8+(18 \times 5)=125$. And if $7+3=10$ be the root; then the cube will be $343+27+(63 \times 10)=1000$. To find the cube of $36$, let us suppose the root $36=30+6$, and we have for the cube required, $27000+216+(540 \times 36)=46656$. 335. But if, on the other hand, the cube be given, namely, $a^3+3a^2b+3ab^2+b^3$, and it be required to find its root, we must premise the following remarks: First, when the cube is arranged according to the powers of one letter, we easily know by the leading term $a^3$, the first term $a$ of the root, since the cube of it is $a^3$; if, therefore, we subtract that cube from the cube proposed, we obtain the remainder, $3a^2b+3ab^2+b^3$, which must furnish the second term of the root. 336. But as we already know, from Art. 333, that the second term is $+b$, we have principally to discover how it may be derived from the above remainder. Now, that remainder may be expressed by two factors, thus, $(3a^2+3ab+b^2) \times (b)$; if, therefore, we divide by $3a^2+3ab+b^2$, we obtain the second part of the root $+b$, which is required. 337. But as this second term is supposed to be unknown, the divisor also is unknown; nevertheless we have the first term of that divisor, which is sufficient: for it is $3a^2$, that is, thrice the square of the first term already found; and by means of this, it is not difficult to find also the other part, $b$, and then to complete the divisor before we perform the division; for this purpose, it will be necessary to join to $3a^2$ thrice the product of the two terms, or $3ab$, and $b^2$, or the square of the second term of the root. 338. Let us apply what we have said to two examples of other given cubes. $a^3+12a^2+48a+64$ ($a+4$ $a^3$ --- $3a^2+12a+16$) $12a^2+48a+64$ $12a^2+48a+64$ --- $0.$