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Question Number 112328 by mohammad17 last updated on 07/Sep/20
find the all root (m^3 −6m+9)=0
$${find}\:{the}\:{all}\:{root}\:\left({m}^{\mathrm{3}} −\mathrm{6}{m}+\mathrm{9}\right)=\mathrm{0} \\ $$
Answered by MJS_new last updated on 07/Sep/20
first try factors of 9: ±1, ±3, ±9  ⇒ m_1 =−3  (m+3)(m^2 −3m+3)=0  ⇒ m_(2, 3) =(3/2)±((√3)/2)i
$$\mathrm{first}\:\mathrm{try}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{9}:\:\pm\mathrm{1},\:\pm\mathrm{3},\:\pm\mathrm{9} \\ $$$$\Rightarrow\:{m}_{\mathrm{1}} =−\mathrm{3} \\ $$$$\left({m}+\mathrm{3}\right)\left({m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{m}_{\mathrm{2},\:\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$
Commented by mohammad17 last updated on 07/Sep/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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