Question Number 228229 by Danmola last updated on 03/Apr/26
$${if}\:{x}^{\mathrm{3}} −\mathrm{8}=\mathrm{0} \\ $$$${Find}\:{all}\:{possible}\:{solutions}\:{for}\:{x} \\ $$
Commented by Danmola last updated on 03/Apr/26
Answered by TonyCWX last updated on 03/Apr/26
![x^3 − 8 = 0 x^3 = 8 x = ((8[cos(0) + isin(0)]))^(1/3) x = { ((2[cos(((0+2(0)π)/3)) + isin(((0+2(0)π)/3))] = 2)),((2[cos(((0+2(1)π)/3)) + isin(((0+2(1)π)/3))] = −1 + i(√3))),((2[cos(((0+2(2)π)/3)) + isin(((0+2(2)π)/3))] = −1 − i(√3))) :}](https://www.tinkutara.com/question/Q228236.png)
$${x}^{\mathrm{3}} \:−\:\mathrm{8}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{3}} \:=\:\mathrm{8} \\ $$$${x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{8}\left[\mathrm{cos}\left(\mathrm{0}\right)\:+\:{i}\mathrm{sin}\left(\mathrm{0}\right)\right]} \\ $$$${x}\:=\:\begin{cases}{\mathrm{2}\left[\mathrm{cos}\left(\frac{\mathrm{0}+\mathrm{2}\left(\mathrm{0}\right)\pi}{\mathrm{3}}\right)\:+\:{i}\mathrm{sin}\left(\frac{\mathrm{0}+\mathrm{2}\left(\mathrm{0}\right)\pi}{\mathrm{3}}\right)\right]\:=\:\mathrm{2}}\\{\mathrm{2}\left[\mathrm{cos}\left(\frac{\mathrm{0}+\mathrm{2}\left(\mathrm{1}\right)\pi}{\mathrm{3}}\right)\:+\:{i}\mathrm{sin}\left(\frac{\mathrm{0}+\mathrm{2}\left(\mathrm{1}\right)\pi}{\mathrm{3}}\right)\right]\:=\:−\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}}\\{\mathrm{2}\left[\mathrm{cos}\left(\frac{\mathrm{0}+\mathrm{2}\left(\mathrm{2}\right)\pi}{\mathrm{3}}\right)\:+\:{i}\mathrm{sin}\left(\frac{\mathrm{0}+\mathrm{2}\left(\mathrm{2}\right)\pi}{\mathrm{3}}\right)\right]\:=\:−\mathrm{1}\:−\:{i}\sqrt{\mathrm{3}}}\end{cases} \\ $$
Answered by fantastic2 last updated on 03/Apr/26
$$\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)=\mathrm{0} \\ $$$${x}−\mathrm{2}=\mathrm{0}\Rightarrow{x}=\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{2}\pm\sqrt{−\mathrm{12}}}{\mathrm{2}}=−\mathrm{1}\pm{i}\sqrt{\mathrm{3}} \\ $$