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x-3-81x-8-1-3-2x-2-4-3-x-2-

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Question Number 198295 by cortano12 last updated on 17/Oct/23
x^3 −((81x−8))^(1/3) = 2x^2 −(4/3)x+2
$$\:\:\:\mathrm{x}^{\mathrm{3}} −\sqrt[{\mathrm{3}}]{\mathrm{81x}−\mathrm{8}}\:=\:\mathrm{2x}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}\mathrm{x}+\mathrm{2}\: \\ $$
Answered by Frix last updated on 17/Oct/23
t=81x−8 t^3 −138t^2 +6348t−531441t^(1/3) −1002754=0 Now it depends on which rule we use: (a) staying in R ⇒ (−z)^(1/3) =−(z^(1/3) ) ★ x=0∨x=1±((2(√6))/3) (b) x∈C ⇒ ∀z∈C: z=re^(iθ) ; r≥0∧−π<θ≤π: z^(1/3) =r^(1/3) e^(i(θ/3)) ⇒ ∀z<0: z^(1/3) =∣z∣e^(i(π/3)) ★ x=1+((2(√6))/3)∨x≈−.517071981±1.43708708i For both cases we can use t=u^3 but we have to check for false solutions. u^9 −138u^6 +6348u^3 −531441u−1002754=0 (u+2)(u^2 −2u−23)(u^6 +27u^4 −92u^3 +729u^2 −1242u+21799)=0 Since 21799 is prime I think we can only approximately solve the last factor.
$${t}=\mathrm{81}{x}−\mathrm{8} \\ $$$${t}^{\mathrm{3}} −\mathrm{138}{t}^{\mathrm{2}} +\mathrm{6348}{t}−\mathrm{531441}{t}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1002754}=\mathrm{0} \\ $$$$\mathrm{Now}\:\mathrm{it}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{which}\:\mathrm{rule}\:\mathrm{we}\:\mathrm{use}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{staying}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow \\ $$$$\left(−{z}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =−\left({z}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$\bigstar\:{x}=\mathrm{0}\vee{x}=\mathrm{1}\pm\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$ \\ $$$$\left(\mathrm{b}\right)\:{x}\in\mathbb{C}\:\Rightarrow \\ $$$$\forall{z}\in\mathbb{C}:\:{z}={r}\mathrm{e}^{\mathrm{i}\theta} ;\:{r}\geqslant\mathrm{0}\wedge−\pi<\theta\leqslant\pi:\:{z}^{\frac{\mathrm{1}}{\mathrm{3}}} ={r}^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{3}}} \\ $$$$\Rightarrow\:\forall{z}<\mathrm{0}:\:{z}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mid{z}\mid\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\bigstar\:{x}=\mathrm{1}+\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}\vee{x}\approx−.\mathrm{517071981}\pm\mathrm{1}.\mathrm{43708708i} \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{both}\:\mathrm{cases}\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:{t}={u}^{\mathrm{3}} \:\mathrm{but}\:\mathrm{we} \\ $$$$\mathrm{have}\:\mathrm{to}\:\mathrm{check}\:\mathrm{for}\:\mathrm{false}\:\mathrm{solutions}. \\ $$$${u}^{\mathrm{9}} −\mathrm{138}{u}^{\mathrm{6}} +\mathrm{6348}{u}^{\mathrm{3}} −\mathrm{531441}{u}−\mathrm{1002754}=\mathrm{0} \\ $$$$\left({u}+\mathrm{2}\right)\left({u}^{\mathrm{2}} −\mathrm{2}{u}−\mathrm{23}\right)\left({u}^{\mathrm{6}} +\mathrm{27}{u}^{\mathrm{4}} −\mathrm{92}{u}^{\mathrm{3}} +\mathrm{729}{u}^{\mathrm{2}} −\mathrm{1242}{u}+\mathrm{21799}\right)=\mathrm{0} \\ $$$$\mathrm{Since}\:\mathrm{21799}\:\mathrm{is}\:\mathrm{prime}\:\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{only} \\ $$$$\mathrm{approximately}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{last}\:\mathrm{factor}. \\ $$

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