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Question Number 127525 by Mathgreat last updated on 30/Dec/20

Commented by Mathgreat last updated on 30/Dec/20

x_1 +x_2 +x_3 +...+x_n = ???

$$\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{3}} +...+\boldsymbol{{x}}_{\boldsymbol{{n}}} =\:??? \\ $$

Commented by Mathgreat last updated on 30/Dec/20

please

$$\boldsymbol{{please}} \\ $$

Answered by MJS_new last updated on 31/Dec/20

x^(7/3) =x^n   n∈N (?)  n=0 ⇒ x^(7/3) =1 ⇒ x=1  n≥1  (1) if we stay in R ⇒ ((−x))^(1/3) =−(x)^(1/3)        ⇒  { ((n=2k; x=0∨x=1)),((n=2k+1; x=−1∨x=0∨x=1)) :}  (2) if we′re in C ⇒ ((−x))^(1/3) =((abs (x)))^(1/3)  e^(i arg(x)/3)        ⇒ x=0∨x=1

$${x}^{\mathrm{7}/\mathrm{3}} ={x}^{{n}} \\ $$$${n}\in\mathbb{N}\:\left(?\right) \\ $$$${n}=\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{7}/\mathrm{3}} =\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$${n}\geqslant\mathrm{1} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{if}\:\mathrm{we}\:\mathrm{stay}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{x}}=−\sqrt[{\mathrm{3}}]{{x}} \\ $$$$\:\:\:\:\:\Rightarrow\:\begin{cases}{{n}=\mathrm{2}{k};\:{x}=\mathrm{0}\vee{x}=\mathrm{1}}\\{{n}=\mathrm{2}{k}+\mathrm{1};\:{x}=−\mathrm{1}\vee{x}=\mathrm{0}\vee{x}=\mathrm{1}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{if}\:\mathrm{we}'\mathrm{re}\:\mathrm{in}\:\mathbb{C}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{x}}=\sqrt[{\mathrm{3}}]{\mathrm{abs}\:\left({x}\right)}\:\mathrm{e}^{\mathrm{i}\:\mathrm{arg}\left({x}\right)/\mathrm{3}} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}=\mathrm{0}\vee{x}=\mathrm{1} \\ $$

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