History of Hindu Mathematics: A Source Book v2.c17.s21.42

v2.c17.s21.42

Put x = mz ± k. Then (x^n ± c)/b = 1/b{ m^n z^n ± n mn-1 zn-1 k &+ (n(n-1))/2 mn-2 zn-2 k² ± / &\dots + n m z (± k)n-1 + (± k)^n ± c } / = 1/b{ m^n z^n ± n mn-1 zn-1 k &+ \dots + n m z (± k)n-1 } / &+ (±)^n ( (k^n ± c)/b ).