History of Hindu Mathematics: A Source Book v2.c17.s21.28
History of Hindu Mathematics: A Source Book v2.c17.s21.28
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v2.c17.s21.28
*Case ii.* c ≠ a cube. Suppose c = bm + n; then find s such that $n + sb = r³.$ Now assume y = bv + r, whence we get $x = b²v³ + 3rv(bv + r) - (m - s),$ as the general solution.