Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com