Any irrational number is the solution of at most one rational depressedcubic equation – that is, an equation of the form x~3+ mx = nfor rationalm, n (n≠0). For if x~* solves both x~3+ m_1x = n_1 and x~3+ m_2x = n2with rational m_1, m_2, n_1and n_2, then subtracting these equations gives(m_1–m_2) x~* = (n_1– n_2),implying x~* is rational unless m_1= m_2andn_1– n_2.Moreover, some algebraic irrational numbers are the solution of norational depressed cubic equation:for example, if 2~(1/2)satisfiesx~3 + mx = n for some m and n with n≠0, we have (2 +m)2~(1/2)= n;hence m and n cannot both be rational.
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