History of Hindu Mathematics: A Source Book v2.c17.s21.26

v2.c17.s21.26

*Case i.* Let c = β³. Then we shall have to find p such that p³ = bq, 3pβ = br. Now assume y = pv + β. Substituting in the equation bx + β³ = y³ we get bx + \beta³ &= (pv + \beta)³ / &= p³v³ + 3pv\beta(pv + \beta) + \beta³, / or \phantom{bx + \beta³} bx &= bqv³ + brv(pv + \beta). / \therefore \phantom{bx + \beta³} x &= qv³ + rv(pv + \beta).