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Question-198644

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Question Number 198644 by cortano12 last updated on 22/Oct/23
Commented by Rasheed.Sindhi last updated on 22/Oct/23
ab=(((8−a+b)/6))^3 or (((8+a−b)/6))^3 ?
$${ab}=\left(\frac{\mathrm{8}−{a}+{b}}{\mathrm{6}}\right)^{\mathrm{3}} \:{or}\:\left(\frac{\mathrm{8}+{a}−{b}}{\mathrm{6}}\right)^{\mathrm{3}} \:? \\ $$
Commented by cortano12 last updated on 23/Oct/23
ab=(((8−a−b)/6))^3
$$\mathrm{ab}=\left(\frac{\mathrm{8}−\mathrm{a}−\mathrm{b}}{\mathrm{6}}\right)^{\mathrm{3}} \\ $$
Answered by Frix last updated on 22/Oct/23
(ab)^(1/3) =((8−a−b)/6) 6(ab)^(1/3) =8−(a+b) 3×2(ab)^(1/3) =2^3 −(a+b) assuming a^(1/3) +b^(1/3) =2 [a^(1/3) +b^(1/3) =c ⇔ a+3(ab)^(1/3) (a^(1/3) +b^(1/3) )_(=c) +b=c^3 ] 3×(a^(1/3) +b^(1/3) )(ab)^(1/3) =2^3 −(a+b) 3a^(2/3) b^(1/3) +3a^(1/3) b^(2/3) =2^3 −(a+b) a+3a^(2/3) b^(1/3) +3a^(1/3) b^(2/3) +b=2^3 (a^(1/3) +b^(1/3) )^3 =2^3 a^(1/3) +b^(1/3) =2 a^(1/3) −b^(1/3) =10 a^(1/3) =6∧b^(1/3) =−4 a=216∧b=−64 a−b=280
$$\left({ab}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{8}−{a}−{b}}{\mathrm{6}} \\ $$$$\mathrm{6}\left({ab}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{8}−\left({a}+{b}\right) \\ $$$$\mathrm{3}×\mathrm{2}\left({ab}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2}^{\mathrm{3}} −\left({a}+{b}\right) \\ $$$$\:\:\:\:\mathrm{assuming}\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2} \\ $$$$\:\:\:\:\:\left[{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}\:\Leftrightarrow\:{a}+\mathrm{3}\left({ab}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \underset{={c}} {\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)}+{b}={c}^{\mathrm{3}} \right] \\ $$$$\mathrm{3}×\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\left({ab}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2}^{\mathrm{3}} −\left({a}+{b}\right) \\ $$$$\mathrm{3}{a}^{\frac{\mathrm{2}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{2}^{\mathrm{3}} −\left({a}+{b}\right) \\ $$$${a}+\mathrm{3}{a}^{\frac{\mathrm{2}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}=\mathrm{2}^{\mathrm{3}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} =\mathrm{2}^{\mathrm{3}} \\ $$$$ \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{10} \\ $$$$ \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{6}\wedge{b}^{\frac{\mathrm{1}}{\mathrm{3}}} =−\mathrm{4} \\ $$$${a}=\mathrm{216}\wedge{b}=−\mathrm{64} \\ $$$${a}−{b}=\mathrm{280} \\ $$
Commented by Frix last updated on 22/Oct/23
I used (−r)^(1/3) =−(r^(1/3) )
$$\mathrm{I}\:\mathrm{used}\:\left(−{r}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =−\left({r}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\: \\ $$
Commented by AST last updated on 23/Oct/23
Okay, you didn′t need to have assumed a^(1/3) +b^(1/3) =2,it followed from the next lines..
$${Okay},\:{you}\:{didn}'{t}\:{need}\:{to}\:{have}\:{assumed} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2},{it}\:{followed}\:{from}\:{the}\:{next}\:{lines}.. \\ $$
Commented by Frix last updated on 23/Oct/23
a^(1/3) −b^(1/3) =216^(1/3) −(−64)^(1/3) =216^(1/3) +64^(1/3) =6+4=10 === ab=216×(−64)=−13824 (((8−a−b)/6))^3 =(((8−216+64)/6))^3 =(−24)^3 =−13824
$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{216}^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(−\mathrm{64}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{216}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{64}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{6}+\mathrm{4}=\mathrm{10} \\ $$$$=== \\ $$$${ab}=\mathrm{216}×\left(−\mathrm{64}\right)=−\mathrm{13824} \\ $$$$\left(\frac{\mathrm{8}−{a}−{b}}{\mathrm{6}}\right)^{\mathrm{3}} =\left(\frac{\mathrm{8}−\mathrm{216}+\mathrm{64}}{\mathrm{6}}\right)^{\mathrm{3}} =\left(−\mathrm{24}\right)^{\mathrm{3}} =−\mathrm{13824} \\ $$
Answered by Frix last updated on 23/Oct/23
Let a=x^3 ∧b=y^3 (1) ⇒ y=x−10 (2) ⇒ x^9 −45x^8 +1125x^7 −18360x^6 +213300x^5 −1798200x^4 +10914048x^3 −45450720x^2 +114307200x−128024064=0 x=6 x^8 −39x^7 +891x^6 −13014x^5 +135216x^4 −986904x^3 +4992624x^2 −15494976x+21337344=0 (x^2 −6x+84)(x^2 −9x+21)(x^2 −12x+84)(x^2 −12x+144)=0 x=3±5(√3)i∨x=(9/2)±((√3)/2)i∨x=6±4(√3)i∨x=6±6(√3)i none of these fit in the first equation because a=x^3 ⇎ a^(1/3) =x x=re^(iα) ⇒ x^3 =r^3 e^(i3α) but 3α=β∧−π<β≤π arg (e^(i3α) ) =(π/2)(1+sign (sin 3α))+mod (−3α, π) ⇒ i.e. x=3+5(√3)i≈9.16515e^(1.23732i) a=x^3 ≈769.873e^(−2.57122i) a^(1/3) ≈9.16515e^(−.857072i) ≠x ⇒ no solution with rules for a, b ∈C 1 solution with rules for a, b ∈R
$$\mathrm{Let}\:{a}={x}^{\mathrm{3}} \wedge{b}={y}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:{y}={x}−\mathrm{10} \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\: \\ $$$${x}^{\mathrm{9}} −\mathrm{45}{x}^{\mathrm{8}} +\mathrm{1125}{x}^{\mathrm{7}} −\mathrm{18360}{x}^{\mathrm{6}} +\mathrm{213300}{x}^{\mathrm{5}} −\mathrm{1798200}{x}^{\mathrm{4}} +\mathrm{10914048}{x}^{\mathrm{3}} −\mathrm{45450720}{x}^{\mathrm{2}} +\mathrm{114307200}{x}−\mathrm{128024064}=\mathrm{0} \\ $$$${x}=\mathrm{6} \\ $$$${x}^{\mathrm{8}} −\mathrm{39}{x}^{\mathrm{7}} +\mathrm{891}{x}^{\mathrm{6}} −\mathrm{13014}{x}^{\mathrm{5}} +\mathrm{135216}{x}^{\mathrm{4}} −\mathrm{986904}{x}^{\mathrm{3}} +\mathrm{4992624}{x}^{\mathrm{2}} −\mathrm{15494976}{x}+\mathrm{21337344}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{84}\right)\left({x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{21}\right)\left({x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{84}\right)\left({x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{144}\right)=\mathrm{0} \\ $$$${x}=\mathrm{3}\pm\mathrm{5}\sqrt{\mathrm{3}}\mathrm{i}\vee{x}=\frac{\mathrm{9}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\vee{x}=\mathrm{6}\pm\mathrm{4}\sqrt{\mathrm{3}}\mathrm{i}\vee{x}=\mathrm{6}\pm\mathrm{6}\sqrt{\mathrm{3}}\mathrm{i} \\ $$$$\mathrm{none}\:\mathrm{of}\:\mathrm{these}\:\mathrm{fit}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{equation}\:\mathrm{because} \\ $$$${a}={x}^{\mathrm{3}} \:\nLeftrightarrow\:{a}^{\frac{\mathrm{1}}{\mathrm{3}}} ={x} \\ $$$${x}={r}\mathrm{e}^{\mathrm{i}\alpha} \:\Rightarrow\:{x}^{\mathrm{3}} ={r}^{\mathrm{3}} \mathrm{e}^{\mathrm{i3}\alpha} \:\mathrm{but}\:\mathrm{3}\alpha=\beta\wedge−\pi<\beta\leqslant\pi \\ $$$$\mathrm{arg}\:\left(\mathrm{e}^{\mathrm{i3}\alpha} \right)\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\mathrm{sign}\:\left(\mathrm{sin}\:\mathrm{3}\alpha\right)\right)+\mathrm{mod}\:\left(−\mathrm{3}\alpha,\:\pi\right) \\ $$$$\Rightarrow\:\mathrm{i}.\mathrm{e}. \\ $$$${x}=\mathrm{3}+\mathrm{5}\sqrt{\mathrm{3}}\mathrm{i}\approx\mathrm{9}.\mathrm{16515e}^{\mathrm{1}.\mathrm{23732i}} \\ $$$${a}={x}^{\mathrm{3}} \approx\mathrm{769}.\mathrm{873e}^{−\mathrm{2}.\mathrm{57122i}} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} \approx\mathrm{9}.\mathrm{16515e}^{−.\mathrm{857072i}} \neq{x} \\ $$$$\Rightarrow \\ $$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{rules}\:\mathrm{for}\:{a},\:{b}\:\in\mathbb{C} \\ $$$$\mathrm{1}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{rules}\:\mathrm{for}\:{a},\:{b}\:\in\mathbb{R} \\ $$

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