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Question Number 26348 by RoachDN last updated on 24/Dec/17

Commented by RoachDN last updated on 24/Dec/17

short trick to solve?

$$\mathrm{short}\:\mathrm{trick}\:\mathrm{to}\:\mathrm{solve}? \\ $$

Answered by $@ty@m last updated on 24/Dec/17

((x^4 +(1/x^4 )+1)/(x^2 +(1/x^2 )+1)) where x=(6/5) =(((x^2 +(1/x^2 ))^2 −1)/(x^2 +(1/x^2 )+1)) =(((x^2 +(1/x^2 )+1)(x^2 +(1/x^2 )−1))/(x^2 +(1/x^2 )+1)) =x^2 +(1/x^2 )−1 =(x−(1/x))^2 +1 =((6/5)−(5/6))^2 +1 =(((36−25)/(30)))^2 +1 =(((11)/(30)))^2 +1 =((121)/(900))+1 =((1021)/(900))

$$\frac{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{1}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}}\:{where}\:{x}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$=\frac{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}} \\ $$$$=\frac{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}} \\ $$$$={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1} \\ $$$$=\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\left(\frac{\mathrm{6}}{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\left(\frac{\mathrm{36}−\mathrm{25}}{\mathrm{30}}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\left(\frac{\mathrm{11}}{\mathrm{30}}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\frac{\mathrm{121}}{\mathrm{900}}+\mathrm{1} \\ $$$$=\frac{\mathrm{1021}}{\mathrm{900}} \\ $$

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