x-1-3-1-x-

Question Number 195017 by mathlove last updated on 22/Jul/23

$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Answered by Rasheed.Sindhi last updated on 22/Jul/23

(x+1)^3 =1 x=? Let x+1=y y^3 −1=0 (y−1)(y^2 +y+1)=0 y−1=0 ∣ y^2 +y+1=0 (x+1)−1=0 ∣ (x+1)^2 +(x+1)+1=0 x=0 ✓ ∣ x^2 +2x+1+x+1+1=0 ∣ x^2 +3x+3=0 x=((−3±(√(9−12)))/2) x=((−3±i(√3))/2)✓

$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$$${Let}\:\:\:\:\:{x}+\mathrm{1}={y} \\ $$$${y}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}^{\mathrm{2}} +{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:{y}−\mathrm{1}=\mathrm{0}\:\mid\:{y}^{\mathrm{2}} +{y}+\mathrm{1}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)−\mathrm{1}=\mathrm{0}\:\mid\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left({x}+\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\checkmark\:\mid\:\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+{x}+\mathrm{1}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{12}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\checkmark \\ $$$$\: \\ $$

Commented by mathlove last updated on 22/Jul/23

$${thanks}\:{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Jul/23

AnOther way x^3 +3x^2 +3x+1=1 x^3 +3x^2 +3x=0 x(x^2 +3x+3)=0 x=0 ∣ x=((−3±(√(9−12)))/2) ∣ x=((−3±i(√3))/2)

$$\mathrm{AnOther}\:\mathrm{way} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}=\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}=\mathrm{0} \\ $$$${x}\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\mid\:{x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{12}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mid\:{x}=\frac{−\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Jul/23

Still another way (x+1)^3 =1 x=? x+1=(1)^(1/3) x+1=1,ω,ω^2 x=0,ω−1,ω^2 −1 x=((−1±i(√3))/2)−1=((−3±i(√3))/2)

$${Still}\:{another}\:{way} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$$${x}+\mathrm{1}=\sqrt[{\mathrm{3}}]{\mathrm{1}}\: \\ $$$${x}+\mathrm{1}=\mathrm{1},\omega,\omega^{\mathrm{2}} \\ $$$${x}=\mathrm{0},\omega−\mathrm{1},\omega^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{1}=\frac{−\mathrm{3}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$

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