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Question Number 180730 by sciencestudent last updated on 16/Nov/22

is it ((125x^3 ))^(1/3) a polynomial?

$${is}\:{it}\:\sqrt[{\mathrm{3}}]{\mathrm{125}{x}^{\mathrm{3}} }\:\:{a}\:{polynomial}? \\ $$

Answered by Frix last updated on 16/Nov/22

No. But for x∈R it is equal to a polynomial: ((125x^3 ))^(1/3) =5x Because in R we define ((−r))^(1/3) =−(r)^(1/3) wherelse for x∈C we have x=re^(iθ) and (x)^(1/3) =(r)^(1/3) e^(i(θ/3)) . in R ((−1))^(1/3) =−(1)^(1/3) =−1 in C ((−1))^(1/3) =(1)^(1/3) e^(i(π/3)) =(1/2)+((√3)/2)i

$$\mathrm{No}.\:\mathrm{But}\:\mathrm{for}\:{x}\in\mathbb{R}\:\mathrm{it}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{a}\:\mathrm{polynomial}: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{125}{x}^{\mathrm{3}} }=\mathrm{5}{x} \\ $$$$\mathrm{Because}\:\mathrm{in}\:\mathbb{R}\:\mathrm{we}\:\mathrm{define}\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}}\:\mathrm{wherelse} \\ $$$$\mathrm{for}\:{x}\in\mathbb{C}\:\mathrm{we}\:\mathrm{have}\:{x}={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{and}\:\sqrt[{\mathrm{3}}]{{x}}=\sqrt[{\mathrm{3}}]{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{3}}} . \\ $$$$\mathrm{in}\:\mathbb{R}\:\sqrt[{\mathrm{3}}]{−\mathrm{1}}=−\sqrt[{\mathrm{3}}]{\mathrm{1}}=−\mathrm{1} \\ $$$$\mathrm{in}\:\mathbb{C}\:\sqrt[{\mathrm{3}}]{−\mathrm{1}}=\sqrt[{\mathrm{3}}]{\mathrm{1}}\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$

Commented by MJS_new last updated on 16/Nov/22

you′re right

$$\mathrm{you}'\mathrm{re}\:\mathrm{right} \\ $$

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