Elements of Algebra

120 ELEMENTS SECT. II. which result is accurate, and confirms the rule. But if we had supposed $a = 1$, $b = 1$, and $c = -1$, we should have found for the cube of $1+1-1$, that is of $1$, $1+3-3+3-6+3+1-3+3-1=1$, which is a still further confirmation of the rule. CHAP. XII. *Of the Expression of Irrational Powers by Infinite Series.* 361. As we have shewn the method of finding any power of the root $a+b$, however great the exponent may be, we are able to express, generally, the power of $a+b$, whose exponent is undetermined; for it is evident that if we represent that exponent by $n$, we shall have by the rule already given (Art. 348 and the following): $(a+b)^n = a^n + \frac{n}{1}a^{n-1}b + \frac{n}{1} \cdot \frac{n-1}{2}a^{n-2}b^2 + \frac{n}{1} \cdot \frac{n-1}{2} \cdot \frac{n-2}{3}a^{n-3}b^3 + \frac{n}{1} \cdot \frac{n-1}{2} \cdot \frac{n-2}{3} \cdot \frac{n-3}{4}a^{n-4}b^4 + \&c.$ 362. If the same power of the root $a-b$ were required, we need only change the signs of the second, fourth, sixth, &c. terms, and should have $(a-b)^n = a^n - \frac{n}{1}a^{n-1}b + \frac{n}{1} \cdot \frac{n-1}{2}a^{n-2}b^2 - \frac{n}{1} \cdot \frac{n-1}{2} \cdot \frac{n-2}{3}a^{n-3}b^3 + \frac{n}{1} \cdot \frac{n-1}{2} \cdot \frac{n-2}{3} \cdot \frac{n-3}{4}a^{n-4}b^4 - \&c.$ 363. These formulas are remarkably useful, since they serve also to express all kinds of radicals; for we have shewn that all irrational quantities may assume the form of powers whose exponents are fractional, and that $\sqrt[2]{a} = a^{\frac{1}{2}}$, $\sqrt[3]{a}=a^{\frac{1}{3}}$, and $\sqrt[4]{a} = a^{\frac{1}{4}}$, &c.: we have, therefore, $\sqrt[2]{(a+b)} = (a+b)^{\frac{1}{2}}$; $\sqrt[3]{(a+b)} = (a+b)^{\frac{1}{3}}$; and $\sqrt[4]{(a+b)} = (a+b)^{\frac{1}{4}}$, &c. Consequently, if we wish to find the square root of $a+b$, we have only to substitute for the exponent $n$ the fraction $\frac{1}{2}$, in the general formula, Art. 361, and we shall have first, for the coefficients,