Solve-x-3-3x-4-0-how-can-i-know-when-to-use-let-x-y-1-y-

Question Number 41798 by Tawa1 last updated on 12/Aug/18

$$\mathrm{Solve}:\:\:\:\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{3x}\:+\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{know}\:\mathrm{when}\:\mathrm{to}\:\mathrm{use}:\:\:\:\:\mathrm{let}\:\mathrm{x}\:=\:\mathrm{y}\:+\:\frac{\mathrm{1}}{\mathrm{y}} \\ $$

Commented by math khazana by abdo last updated on 13/Aug/18

let x =u+v ⇒(u+v)^3 −3(u+v)+4=0 ⇒ u^3 +v^3 +3uv(u+v)−3(u+v) +4 =0 ⇒ u^3 +v^3 +4 +(u+v)(3uv−3)=0 ⇒ u^3 +v^3 =−4 and uv=1 ⇒u^3 +v^3 =−4 and u^3 .v^3 =1 ⇒u^3 and v^3 are solution of the equation X^2 +4X +1 =0 Δ^′ =2^2 −1 =3 ⇒X_1 =−2+(√3) and X_2 =−2−(√3) ⇒ ⇒ u =^3 (√(−2+(√3))) and v =^3 (√(−2−(√3))) ⇒ x =^3 (√(−2+(√3))) +^3 (√(−2−(√3)))

$${let}\:{x}\:={u}+{v}\:\Rightarrow\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}\left({u}+{v}\right)+\mathrm{4}=\mathrm{0}\:\Rightarrow \\ $$$${u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} \:+\mathrm{3}{uv}\left({u}+{v}\right)−\mathrm{3}\left({u}+{v}\right)\:+\mathrm{4}\:=\mathrm{0}\:\Rightarrow \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:+\mathrm{4}\:\:+\left({u}+{v}\right)\left(\mathrm{3}{uv}−\mathrm{3}\right)=\mathrm{0}\:\Rightarrow \\ $$$${u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} \:=−\mathrm{4}\:\:{and}\:{uv}=\mathrm{1}\:\Rightarrow{u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} \:=−\mathrm{4}\:{and} \\ $$$${u}^{\mathrm{3}} .{v}^{\mathrm{3}} \:=\mathrm{1}\:\Rightarrow{u}^{\mathrm{3}} \:{and}\:{v}^{\mathrm{3}} \:{are}\:{solution}\:{of}\:{the}\:{equation} \\ $$$${X}^{\mathrm{2}} \:+\mathrm{4}{X}\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta^{'} \:=\mathrm{2}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{3}\:\Rightarrow{X}_{\mathrm{1}} =−\mathrm{2}+\sqrt{\mathrm{3}}\:\:{and}\:{X}_{\mathrm{2}} =−\mathrm{2}−\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$\Rightarrow\:{u}\:=^{\mathrm{3}} \sqrt{−\mathrm{2}+\sqrt{\mathrm{3}}}\:\:{and}\:\:{v}\:=^{\mathrm{3}} \sqrt{−\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:{x}\:=^{\mathrm{3}} \sqrt{−\mathrm{2}+\sqrt{\mathrm{3}}}\:+^{\mathrm{3}} \sqrt{−\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$

Commented by Tawa1 last updated on 13/Aug/18