(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...


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

$\frac{1}{t + \sqrt{u} + \sqrt{v}} =$ $= \frac{(t - \sqrt{u} + \sqrt{v}) (t + \sqrt{u} - \sqrt{v}) (t - \sqrt{u} - \sqrt{v})}{(t + \sqrt{u} + \sqrt{v}) (t - \sqrt{u} + \sqrt{v}) (t + \sqrt{u} - \sqrt{v}) (t - \sqrt{u} - \sqrt{v})} =$ $= \frac{t^{3} - (\sqrt{u} + \sqrt{v}) t^{2} - {(\sqrt{u} - \sqrt{v})}^{2} t + (u - v) (\sqrt{u} - \sqrt{v})}{t^{4} - 2 (u + v) t^{2} + {(u - v)}^{2}}$ $\frac{1}{t + \sqrt[3]{u} + \sqrt[3]{v}} =$ $\| \begin{matrix} α = - \frac{1}{2} + \frac{\sqrt{3}}{2} i; β = - \frac{1}{2} - \frac{\sqrt{3}}{2} i \Rightarrow \\ \Rightarrow α^{2} = β; β^{2} = α; α + β = - 1; α β = 1 \end{matrix} \|$ $= \frac{(t + α \sqrt[3]{u} + β \sqrt[3]{v}) (t + β \sqrt[3]{u} + α \sqrt[3]{v})}{(t + \sqrt[3]{u} + \sqrt[3]{v}) (t + α \sqrt[3]{u} + β \sqrt[3]{v}) (t + β \sqrt[3]{u} + α \sqrt[3]{v})} =$ $= \frac{t^{2} - (\sqrt[3]{u} + \sqrt[3]{v}) t + \sqrt[3]{u^{2}} - \sqrt[3]{u v} + \sqrt[3]{v^{2}}}{t^{3} + u + v - 3 \sqrt[3]{u v} t} =$ $\| \begin{matrix} a = t^{3} + u + v; b = 27 {u v t}^{3} \\ \frac{1}{a - \sqrt[3]{b}} = \frac{(α a - β \sqrt[3]{b}) (β a - α \sqrt[3]{b})}{(a - \sqrt[3]{b}) (α a - β \sqrt[3]{b}) (β a - α \sqrt[3]{b})} = \\ = \frac{a^{2} + a \sqrt[3]{b} + \sqrt[3]{b^{2}}}{a^{3} - b} \end{matrix} \|$ $= \frac{N}{t^{9} + 3 (u + v) t^{6} + 3 (u^{2} - 7 u v + v^{2}) t^{3} + {(u + v)}^{3}}$ $N =$ $= t^{8} -$ $- (\sqrt[3]{u} + \sqrt[3]{v}) t^{7} +$ $+ {(\sqrt[3]{u} + \sqrt[3]{v})}^{2} t^{6} +$ $+ (\sqrt[3]{u} + \sqrt[3]{v}) (\sqrt[3]{u} - 2 \sqrt[3]{v}) (2 \sqrt[3]{u} - \sqrt[3]{v}) t^{5} -$ $- {(\sqrt[3]{u} + \sqrt[3]{v})}^{2} (\sqrt[3]{u} - 2 \sqrt[3]{v}) (2 \sqrt[3]{u} - \sqrt[3]{v}) t^{4} +$ $+ {(\sqrt[3]{u} + \sqrt[3]{v})}^{3} (\sqrt[3]{u} - 2 \sqrt[3]{v}) (2 \sqrt[3]{u} - \sqrt[3]{v}) t^{3} +$ $+ {(\sqrt[3]{u^{2}} - \sqrt[3]{u v} + \sqrt[3]{v^{2}})}^{3} t^{2} -$ $- (\sqrt[3]{u} + \sqrt[3]{v}) {(\sqrt[3]{u^{2}} - \sqrt[3]{u v} + \sqrt[3]{v^{2}})}^{3} t +$ $+ {(\sqrt[3]{u} + \sqrt[3]{v})}^{2} {(\sqrt[3]{u^{2}} - \sqrt[3]{u v} + \sqrt[3]{v^{2}})}^{3}$ $I think {there}^{'} s no easier way . . .$
Commented by MJS last updated on 08/Oct/19

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