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27x-3-1-0-




Question Number 58091 by smiak8742 last updated on 17/Apr/19
27x^(3−1=0)
$$\mathrm{27}{x}^{\mathrm{3}−\mathrm{1}=\mathrm{0}} \\ $$
Answered by MJS last updated on 17/Apr/19
27x^3 =1  x^3 =(1/(27))  x_1 =(1/3)  x_2 =−(1/6)+((√3)/6)i  x_3 =−(1/6)−((√3)/6)i
$$\mathrm{27}{x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{27}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{6}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{i} \\ $$$${x}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{i} \\ $$
Commented by MJS last updated on 18/Apr/19
x^3 =r  ⇒ x_1 =(r)^(1/3) ; x_2 =(r)^(1/3) (−(1/2)−((√3)/2)i); x_3 =(r)^(1/3) (−(1/2)+((√3)/2)i)
$${x}^{\mathrm{3}} ={r} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{{r}};\:{x}_{\mathrm{2}} =\sqrt[{\mathrm{3}}]{{r}}\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right);\:{x}_{\mathrm{3}} =\sqrt[{\mathrm{3}}]{{r}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$
Answered by Kunal12588 last updated on 17/Apr/19
27x^3 −1=0  ⇒(3x)^3 −(1)^3 =0  ⇒(3x−1)(9x^2 +3x+1)=0  ⇒x=(1/3) or 9x^2 +3x+1=0⇒x=((−3±(√(9−36)))/(18))  ⇒x=(1/3) or x=((i(√3)−1)/6) or x=((−1−i(√3))/6)
$$\mathrm{27}{x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3}{x}\right)^{\mathrm{3}} −\left(\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3}{x}−\mathrm{1}\right)\left(\mathrm{9}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:\mathrm{9}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}=\mathrm{0}\Rightarrow{x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{36}}}{\mathrm{18}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:{x}=\frac{{i}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{6}}\:{or}\:{x}=\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$ \\ $$

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