Elements of Algebra

CHAP. VII. OF ALGEBRA. 103 $a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$ ($a^3-3a^2b+3ab^2-b^3$ $a^6$ --- $2a^3-3a^2b)$ $-6a^5b+15a^4b^2$ $-6a^5b+9a^4b^2$ --- $2a^3-6a^2b+3ab^2)$ $6a^4b^2-20a^3b^3+15a^2b^4$ $6a^4b^2-18a^3b^3+9a^2b^4$ --- $2a^3-6a^2b+6ab^2-b^3)$ $-2a^3b^3+6a^2b^4-6ab^5+b^6$ $-2a^3b^3+6a^2b^4-6ab^5+b^6$ --- $0$. 324. We easily deduce from the rule which we have explained, the method which is taught in books of arithmetic for the extraction of the square root, as will appear from the following examples in numbers: $529$ ($23$ $4$ --- $43$) $129$ $129$ --- $0$. $4096$ ($64$ $36$ --- $124$) $496$ $496$ --- $0$. $15625$ ($125$ $1$ --- $22$) $56$ $44$ --- $245$) $1225$ $1225$ --- $0$. $2304$ ($48$ $16$ --- $88$) $704$ $704$ --- $0$. $9604$ ($98$ $81$ --- $188$) $1504$ $1504$ --- $0$. $998001$ ($999$ $81$ --- $189$) $1880$ $1701$ --- $1989$) $17901$ $17901$ --- $0$. 325. But when there is a remainder after all the figures have been used, it is a proof that the number proposed is