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Question Number 32609 by naka3546 last updated on 01/Apr/18

1. Suppose  that  a, b, and  c  are  real  numbers  such  that  a < b < c  and   a^3  − 3a + 1  =  b^3  − 3b + 1  =  c^3  − 3c + 1 =  0 .      Then   (1/(a^2  + b)) + (1/(b^2  + c)) + (1/(c^2  + a))    can be  written  as   (p/q)   for  relatively  prime  of  positive  integers  p  and   q.   Find   100p + q

$$\mathrm{1}.\:{Suppose}\:\:{that}\:\:{a},\:{b},\:{and}\:\:{c}\:\:{are}\:\:{real}\:\:{numbers}\:\:{such}\:\:{that}\:\:{a}\:<\:{b}\:<\:{c}\:\:{and}\:\:\:{a}^{\mathrm{3}} \:−\:\mathrm{3}{a}\:+\:\mathrm{1}\:\:=\:\:{b}^{\mathrm{3}} \:−\:\mathrm{3}{b}\:+\:\mathrm{1}\:\:=\:\:{c}^{\mathrm{3}} \:−\:\mathrm{3}{c}\:+\:\mathrm{1}\:=\:\:\mathrm{0}\:. \\ $$ $$\:\:\:\:{Then}\:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \:+\:{b}}\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} \:+\:{c}}\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{2}} \:+\:{a}}\:\:\:\:{can}\:{be}\:\:{written}\:\:{as}\:\:\:\frac{{p}}{{q}}\:\:\:{for}\:\:{relatively}\:\:{prime}\:\:{of}\:\:{positive}\:\:{integers}\:\:\boldsymbol{{p}}\:\:{and}\:\:\:\boldsymbol{{q}}.\:\:\:{Find}\:\:\:\mathrm{100}{p}\:+\:{q} \\ $$

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