Question Number 166695 by amin96 last updated on 25/Feb/22
Answered by som(math1967) last updated on 25/Feb/22
![(x^(x+(1/x)) +1+1+x^(x+(1/x)) ) =x^3 +(1/x^3 ) +2 [x+(1/x)=3] =(x+(1/x))^3 −3(x+(1/x))+2 =3^3 −9+2=20](Q166696.png)
$$\left({x}^{{x}+\frac{\mathrm{1}}{{x}}} +\mathrm{1}+\mathrm{1}+{x}^{{x}+\frac{\mathrm{1}}{{x}}} \right) \\ $$$$={x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:+\mathrm{2}\:\:\:\left[{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\right] \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{2} \\ $$$$=\mathrm{3}^{\mathrm{3}} −\mathrm{9}+\mathrm{2}=\mathrm{20} \\ $$
Answered by MJS_new last updated on 25/Feb/22
$$\left({x}^{{x}} +\left(\mathrm{1}/{x}\right)^{\mathrm{1}/{x}} \right)\left({x}^{\mathrm{1}/{x}} +\left(\mathrm{1}/{x}\right)^{{x}} \right)= \\ $$$$={x}^{{x}+\frac{\mathrm{1}}{{x}}} +\frac{\mathrm{1}}{{x}^{{x}+\frac{\mathrm{1}}{{x}}} }+\mathrm{2}={x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{2}= \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{2}\right)=\mathrm{2}^{\mathrm{2}} ×\mathrm{5}=\mathrm{20} \\ $$