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Question Number 70780 by MJS last updated on 08/Oct/19

(1/(t+(√u)+(√v)))= =(((t−(√u)+(√v))(t+(√u)−(√v))(t−(√u)−(√v)))/((t+(√u)+(√v))(t−(√u)+(√v))(t+(√u)−(√v))(t−(√u)−(√v))))= =((t^3 −((√u)+(√v))t^2 −((√u)−(√v))^2 t+(u−v)((√u)−(√v)))/(t^4 −2(u+v)t^2 +(u−v)^2 )) (1/(t+(u)^(1/3) +(v)^(1/3) ))= determinant (((α=−(1/2)+((√3)/2)i; β=−(1/2)−((√3)/2)i ⇒)),((⇒ α^2 =β; β^2 =α; α+β=−1; αβ=1))) =(((t+α(u)^(1/3) +β(v)^(1/3) )(t+β(u)^(1/3) +α(v)^(1/3) ))/((t+(u)^(1/3) +(v)^(1/3) )(t+α(u)^(1/3) +β(v)^(1/3) )(t+β(u)^(1/3) +α(v)^(1/3) )))= =((t^2 −((u)^(1/3) +(v)^(1/3) )t+(u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )/(t^3 +u+v−3((uv))^(1/3) t))= determinant (((a=t^3 +u+v; b=27uvt^3 )),(((1/(a−(b)^(1/3) ))=(((αa−β(b)^(1/3) )(βa−α(b)^(1/3) ))/((a−(b)^(1/3) )(αa−β(b)^(1/3) )(βa−α(b)^(1/3) )))=)),((=((a^2 +a(b)^(1/3) +(b^2 )^(1/3) )/(a^3 −b))))) =(N/(t^9 +3(u+v)t^6 +3(u^2 −7uv+v^2 )t^3 +(u+v)^3 )) N= =t^8 − −((u)^(1/3) +(v)^(1/3) )t^7 + +((u)^(1/3) +(v)^(1/3) )^2 t^6 + +((u)^(1/3) +(v)^(1/3) )((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^5 − −((u)^(1/3) +(v)^(1/3) )^2 ((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^4 + +((u)^(1/3) +(v)^(1/3) )^3 ((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^3 + +((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 t^2 − −((u)^(1/3) +(v)^(1/3) )((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 t+ +((u)^(1/3) +(v)^(1/3) )^2 ((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 I think there′s no easier way...

$$\frac{\mathrm{1}}{{t}+\sqrt{{u}}+\sqrt{{v}}}= \\ $$$$=\frac{\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}{\left({t}+\sqrt{{u}}+\sqrt{{v}}\right)\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}= \\ $$$$=\frac{{t}^{\mathrm{3}} −\left(\sqrt{{u}}+\sqrt{{v}}\right){t}^{\mathrm{2}} −\left(\sqrt{{u}}−\sqrt{{v}}\right)^{\mathrm{2}} {t}+\left({u}−{v}\right)\left(\sqrt{{u}}−\sqrt{{v}}\right)}{{t}^{\mathrm{4}} −\mathrm{2}\left({u}+{v}\right){t}^{\mathrm{2}} +\left({u}−{v}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{t}+\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}}= \\ $$$$\:\:\:\:\:\begin{vmatrix}{\alpha=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i};\:\beta=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow}\\{\Rightarrow\:\alpha^{\mathrm{2}} =\beta;\:\beta^{\mathrm{2}} =\alpha;\:\alpha+\beta=−\mathrm{1};\:\alpha\beta=\mathrm{1}}\end{vmatrix} \\ $$$$=\frac{\left({t}+\alpha\sqrt[{\mathrm{3}}]{{u}}+\beta\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\beta\sqrt[{\mathrm{3}}]{{u}}+\alpha\sqrt[{\mathrm{3}}]{{v}}\right)}{\left({t}+\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\alpha\sqrt[{\mathrm{3}}]{{u}}+\beta\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\beta\sqrt[{\mathrm{3}}]{{u}}+\alpha\sqrt[{\mathrm{3}}]{{v}}\right)}= \\ $$$$=\frac{{t}^{\mathrm{2}} −\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right){t}+\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }}{{t}^{\mathrm{3}} +{u}+{v}−\mathrm{3}\sqrt[{\mathrm{3}}]{{uv}}\:{t}}= \\ $$$$\:\:\:\:\:\begin{vmatrix}{{a}={t}^{\mathrm{3}} +{u}+{v};\:{b}=\mathrm{27}{uvt}^{\mathrm{3}} }\\{\frac{\mathrm{1}}{{a}−\sqrt[{\mathrm{3}}]{{b}}}=\frac{\left(\alpha{a}−\beta\sqrt[{\mathrm{3}}]{{b}}\right)\left(\beta{a}−\alpha\sqrt[{\mathrm{3}}]{{b}}\right)}{\left({a}−\sqrt[{\mathrm{3}}]{{b}}\right)\left(\alpha{a}−\beta\sqrt[{\mathrm{3}}]{{b}}\right)\left(\beta{a}−\alpha\sqrt[{\mathrm{3}}]{{b}}\right)}=}\\{=\frac{{a}^{\mathrm{2}} +{a}\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }}{{a}^{\mathrm{3}} −{b}}}\end{vmatrix} \\ $$$$=\frac{{N}}{{t}^{\mathrm{9}} +\mathrm{3}\left({u}+{v}\right){t}^{\mathrm{6}} +\mathrm{3}\left({u}^{\mathrm{2}} −\mathrm{7}{uv}+{v}^{\mathrm{2}} \right){t}^{\mathrm{3}} +\left({u}+{v}\right)^{\mathrm{3}} } \\ $$$${N}= \\ $$$$={t}^{\mathrm{8}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{7}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} {t}^{\mathrm{6}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{5}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{4}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{3}} \left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{3}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} {t}^{\mathrm{2}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} {t}+ \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{easier}\:\mathrm{way}... \\ $$

Commented by MJS last updated on 08/Oct/19

this had been requested (but deleted) (1/(1+2(3)^(1/3) +3(2)^(1/3) ))= =((−8208((36))^(1/3) +2988((18))^(1/3) −378((12))^(1/3) +23020(9)^(1/3) −36024(6)^(1/3) +54225(4)^(1/3) +13114(3)^(1/3) −1659(2)^(1/3) −5423)/(458047))

$$\mathrm{this}\:\mathrm{had}\:\mathrm{been}\:\mathrm{requested}\:\left(\mathrm{but}\:\mathrm{deleted}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}}= \\ $$$$=\frac{−\mathrm{8208}\sqrt[{\mathrm{3}}]{\mathrm{36}}+\mathrm{2988}\sqrt[{\mathrm{3}}]{\mathrm{18}}−\mathrm{378}\sqrt[{\mathrm{3}}]{\mathrm{12}}+\mathrm{23020}\sqrt[{\mathrm{3}}]{\mathrm{9}}−\mathrm{36024}\sqrt[{\mathrm{3}}]{\mathrm{6}}+\mathrm{54225}\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{13114}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\mathrm{1659}\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{5423}}{\mathrm{458047}} \\ $$

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