Good one!

111/x + 111/x² - 11/x = 11 Bringing all terms to LHS: 111/x + 111/x² - 11/x³ - 11 = 0 Substituting a = 1/x, a² = 1/x², a³ = 1/x³: 111a + 111a² - 11a³ - 11 = 0 111a(1 + a) - 11(1 + a³) = 0 Using 1 + b³ = (1 + b)(1 - b + b²): 111a(1 + a) - 11(1 + a)(1 - a + a²) = 0 (1 + a)(111a - 11(1 - a + a²)) = 0 (1 + a)(111a - 11 + 11a - 11a²) = 0 (1 + a)(-1)(11a² - 122a + 11) = 0 Negating both sides: (1 + a)(11a² - 121a - a + 11) = 0 (1 + a)(11a(a - 11) - 1(a - 11)) = 0 (1 + a)(11a - 1)(a - 11) = 0 Case 1: 1 + a = 0 a = -1. Case 2: 11a - 1 = 0 a = 1/11. Case 3: a - 11 = 0 a = 11. But a = 1/x → x = 1/a: From Case 1: x = 1/(-1) x = -1. From Case 2: x = 1/(1/11) x = 11. From Case 3: x = 1/11. Thus x = {-1, 1/11, 11}.

Another way: Dividing by -11, the original equation is 1/x^3 - (111/11)/x^2 - (111/11)/x + 1=0. Multiplying by x^3, we obtain x^3 - (111/11)x^2 - (111/11)x + 1=0, which is the same equation substituting x by 1/x. Therefore, if r is a root, then 1/r is also a root. As the equation must have 3 roots (counting multiplicity), one of them must be 1 or -1 which are the (complex) numbers which are equal to their reciprocals. By inspection, -1 is root. We have then other roots r and 1/r. We know the sum of the roots must be 111/11 (Girard relation), therefore r + 1/r - 1 = 111/11. Rearranging terms, r + 1/r = 111/11 + 1 = 10 + 1/11 + 1 = 11 + 1/11, therefore r = 11 and the other root must be 1/11 (the same result can be obtained solving a quadratic equation in r).

@rashel1

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thanks