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Question Number 211201 by efronzo1 last updated on 31/Aug/24
((x−1))^(1/3) + ((x−2))^(1/3) −((2x−3))^(1/3) = 0 x=?
$$\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\mathrm{2x}−\mathrm{3}}\:=\:\mathrm{0} \\ $$$$\:\:\:\mathrm{x}=? \\ $$
Answered by Rasheed.Sindhi last updated on 31/Aug/24
((x−1))^(1/3) + ((x−2))^(1/3) −((2x−3))^(1/3) = 0 ∵−((2x−3))^(1/3) =((3−2x))^(1/3) ∴ ((x−1))^(1/3) + ((x−2))^(1/3) +((3−2x))^(1/3) = 0 determinant (((a+b+c=0⇒a^3 +b^3 +c^3 −3abc=0))) ( ((x−1))^(1/3) )^3 +( ((x−2))^(1/3) )^3 +(((3−2x))^(1/3) )^3 −3(((x−1)(x−2)(3−2x)))^(1/3) =0 (x−1)+(x−2)+(3−2x)−3(((x−1)(x−2)(3−2x)))^(1/3) =0 (2x−3)+(3−2x)=3(((x−1)(x−2)(3−2x)))^(1/3) 0=27(x−1)(x−2)(3−2x) (x−1)(x−2)(3−2x)=0 x=1,2,(3/2)
$$\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\mathrm{2x}−\mathrm{3}}\:=\:\mathrm{0} \\ $$$$\because−\sqrt[{\mathrm{3}}]{\mathrm{2x}−\mathrm{3}}\:=\sqrt[{\mathrm{3}}]{\mathrm{3}−\mathrm{2x}} \\ $$$$\:\:\therefore\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{3}−\mathrm{2x}}\:=\:\mathrm{0} \\ $$$$\begin{array}{|c|}{{a}+{b}+{c}=\mathrm{0}\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{0}}\\\hline\end{array} \\ $$$$\:\:\left(\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}\:\right)^{\mathrm{3}} +\left(\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}}\:\right)^{\mathrm{3}} +\left(\sqrt[{\mathrm{3}}]{\mathrm{3}−\mathrm{2x}}\:\right)^{\mathrm{3}} −\mathrm{3}\sqrt[{\mathrm{3}}]{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{3}−\mathrm{2x}\right)}\:=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)+\left(\mathrm{x}−\mathrm{2}\right)+\left(\mathrm{3}−\mathrm{2x}\right)−\mathrm{3}\sqrt[{\mathrm{3}}]{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{3}−\mathrm{2x}\right)}\:=\mathrm{0}\: \\ $$$$\left(\mathrm{2x}−\mathrm{3}\right)+\left(\mathrm{3}−\mathrm{2x}\right)=\mathrm{3}\sqrt[{\mathrm{3}}]{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{3}−\mathrm{2x}\right)} \\ $$$$\mathrm{0}=\mathrm{27}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{3}−\mathrm{2x}\right) \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{3}−\mathrm{2x}\right)=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{1},\mathrm{2},\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by gri_bor last updated on 31/Aug/24
((x−1))^(1/3) +((x−2))^(1/3) =((2x−3))^(1/3) ∣ ()^3 x−1+3×(((x−1)^2 (x−2)))^(1/3) +3×(((x−1)(x−2)^2 ))^(1/3) +x−2=2x−3 3×(((x−1)^2 (x−2)))^(1/3) +3×(((x−1)(x−2)^2 ))^(1/3) =0 3×(((x−1)^2 (x−2)))^(1/3) =−3×(((x−1)(x−2)^2 ))^(1/3) ∣ /3 (((x−1)^2 (x−2)))^(1/3) =−(((x−1)(x−2)^2 ))^(1/3) ∣ ()^3 (x−1)^2 (x−2)=−(x−1)(x−2)^2 ∣ /(x−1)(x−2) x−1=−(x−2) x−1=2−x 2x=3 x=(3/2)=1,5
$$\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}=\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{3}}\:\mid\:\left(\right)^{\mathrm{3}} \\ $$$${x}−\mathrm{1}+\mathrm{3}×\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)}+\mathrm{3}×\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+{x}−\mathrm{2}=\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{3}×\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)}+\mathrm{3}×\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{3}×\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)}=−\mathrm{3}×\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:\mid\:/\mathrm{3} \\ $$$$\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)}=−\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:\mid\:\left(\right)^{\mathrm{3}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)=−\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:\mid\:/\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right) \\ $$$${x}−\mathrm{1}=−\left({x}−\mathrm{2}\right) \\ $$$${x}−\mathrm{1}=\mathrm{2}−{x} \\ $$$$\mathrm{2}{x}=\mathrm{3} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1},\mathrm{5} \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 31/Aug/24
Line 6: you divide by (x−1)(x−2) but what if (x−1)(x−2)=0? You lost 2 solutions there... We are never allowed to simply divide by any factor containing a variable. [5x=3x ⇒ x=0 but ((5x)/x)=((3x)/x) ⇒ 5=3] The correct path is: (x−1)^2 (x−2)=−(x−1)(x−2)^2 (x−1)^2 (x−2)+(x−1)(x−2)^2 =0 (x−1)(x−2)((x−1)+(x−2))=0 (x−1_(x=1) )(x−2_(x=2) )(2x−3_(x=(3/2)) )=0
$$\mathrm{Line}\:\mathrm{6}:\:\mathrm{you}\:\mathrm{divide}\:\mathrm{by}\:\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\:\mathrm{but} \\ $$$$\mathrm{what}\:\mathrm{if}\:\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{0}? \\ $$$$\mathrm{You}\:\mathrm{lost}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{there}… \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{never}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{simply}\:\mathrm{divide}\:\mathrm{by} \\ $$$$\mathrm{any}\:\mathrm{factor}\:\mathrm{containing}\:\mathrm{a}\:\mathrm{variable}. \\ $$$$\left[\mathrm{5}{x}=\mathrm{3}{x}\:\Rightarrow\:{x}=\mathrm{0}\:\mathrm{but}\:\frac{\mathrm{5}{x}}{{x}}=\frac{\mathrm{3}{x}}{{x}}\:\Rightarrow\:\mathrm{5}=\mathrm{3}\right] \\ $$$$\mathrm{The}\:\mathrm{correct}\:\mathrm{path}\:\mathrm{is}: \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)=−\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{2}\right)+\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left(\left({x}−\mathrm{1}\right)+\left({x}−\mathrm{2}\right)\right)=\mathrm{0} \\ $$$$\left(\underset{{x}=\mathrm{1}} {\underbrace{{x}−\mathrm{1}}}\right)\left(\underset{{x}=\mathrm{2}} {\underbrace{{x}−\mathrm{2}}}\right)\left(\underset{{x}=\frac{\mathrm{3}}{\mathrm{2}}} {\underbrace{\mathrm{2}{x}−\mathrm{3}}}\right)=\mathrm{0} \\ $$
Answered by Frix last updated on 31/Aug/24
All calculations stay in R ⇒ ((−r))^(1/3) =−(r)^(1/3) [in C: r∈R ⇒ ((−r))^(1/3) =((re^(iπ) ))^(1/3) =(r)^(1/3) e^(i(π/3)) ≠−(r)^(1/3) ] (x−1)^(1/3) +(x−2)^(1/3) =(2x−3)^(1/3) [(a^(1/3) +b^(1/3) =c^(1/3) )^3 ⇒ a+b+3a^(1/3) b^(1/3) (a^(1/3) +b^(1/3) _(=c^(1/3) ) )=c] ⇒ (x−1)+(x−2)+3(x−1)^(1/3) (x−2)^(1/3) (2x−3)^(1/3) =2x−3 (x−1)^(1/3) (x−2)^(1/3) (2x−3)^(1/3) =0 x=1∨x=2∨x=(3/2) [with ((−r))^(1/3) =(r)^(1/3) e^(i(π/3)) : { ((x=1: 0+((−1))^(1/3) =((−1))^(1/3) true)),((x=(3/2): ((1/2))^(1/3) +((−(1/2)))^(1/3) =0 false)),((x=2: (1)^(1/3) +0=(1)^(1/3) true)) :}]
$$\mathrm{All}\:\mathrm{calculations}\:\mathrm{stay}\:\mathrm{in}\:\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\:\:\:\:\:\left[\mathrm{in}\:\mathbb{C}:\:{r}\in\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{r}}=\sqrt[{\mathrm{3}}]{{r}\mathrm{e}^{\mathrm{i}\pi} }=\sqrt[{\mathrm{3}}]{{r}}\:\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} \neq−\sqrt[{\mathrm{3}}]{{r}}\right] \\ $$$$\left({x}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\left({x}−\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\left(\mathrm{2}{x}−\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\left[\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} \Rightarrow\:{a}+{b}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\underset{={c}^{\frac{\mathrm{1}}{\mathrm{3}}} } {\underbrace{{a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} }}\right)={c}\right] \\ $$$$\Rightarrow \\ $$$$\left({x}−\mathrm{1}\right)+\left({x}−\mathrm{2}\right)+\mathrm{3}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left({x}−\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{2}{x}−\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{2}{x}−\mathrm{3} \\ $$$$\left({x}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left({x}−\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{2}{x}−\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{0} \\ $$$${x}=\mathrm{1}\vee{x}=\mathrm{2}\vee{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\left[\mathrm{with}\:\sqrt[{\mathrm{3}}]{−{r}}=\sqrt[{\mathrm{3}}]{{r}}\:\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}} :\:\begin{cases}{{x}=\mathrm{1}:\:\mathrm{0}+\sqrt[{\mathrm{3}}]{−\mathrm{1}}=\sqrt[{\mathrm{3}}]{−\mathrm{1}}\:\mathrm{true}}\\{{x}=\frac{\mathrm{3}}{\mathrm{2}}:\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{0}\:\mathrm{false}}\\{{x}=\mathrm{2}:\:\sqrt[{\mathrm{3}}]{\mathrm{1}}+\mathrm{0}=\sqrt[{\mathrm{3}}]{\mathrm{1}}\:\mathrm{true}}\end{cases}\right] \\ $$
Answered by mr W last updated on 31/Aug/24
(((x−(3/2))+(1/2)))^(1/3) +(((x−(3/2))−(1/2)))^(1/3) −((2(x−(3/2))))^(1/3) =0 let t=x−(3/2) ((t+(1/2)))^(1/3) +((t−(1/2)))^(1/3) −((2t))^(1/3) =0 ⇒t=0, −(1/2), (1/2) ⇒x=(3/2), 1, 2 ✓
$$\sqrt[{\mathrm{3}}]{\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}}+\sqrt[{\mathrm{3}}]{\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)}=\mathrm{0} \\ $$$${let}\:{t}={x}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sqrt[{\mathrm{3}}]{{t}+\frac{\mathrm{1}}{\mathrm{2}}}+\sqrt[{\mathrm{3}}]{{t}−\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}{t}}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{0},\:−\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{1},\:\mathrm{2}\:\checkmark \\ $$
Commented by mr W last updated on 31/Aug/24
we see this way that the solutions are always (3/2), 1, 2, even when the equation is ((x−1))^(1/7) +((x−2))^(1/7) −((2x−3))^(1/7) =0 or ((x−1))^(1/7) +((x−2))^(1/7) −((2x−3))^(1/(11)) =0 etc.
$${we}\:{see}\:{this}\:{way}\:{that}\:{the}\:{solutions}\: \\ $$$${are}\:{always}\:\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{1},\:\mathrm{2},\:{even}\:{when}\:{the}\: \\ $$$${equation}\:{is} \\ $$$$\sqrt[{\mathrm{7}}]{{x}−\mathrm{1}}+\sqrt[{\mathrm{7}}]{{x}−\mathrm{2}}−\sqrt[{\mathrm{7}}]{\mathrm{2}{x}−\mathrm{3}}=\mathrm{0}\:{or} \\ $$$$\sqrt[{\mathrm{7}}]{{x}−\mathrm{1}}+\sqrt[{\mathrm{7}}]{{x}−\mathrm{2}}−\sqrt[{\mathrm{11}}]{\mathrm{2}{x}−\mathrm{3}}=\mathrm{0} \\ $$$${etc}. \\ $$
Answered by RojaTaniya last updated on 31/Aug/24
((x−1))^(1/3) + ((x−2))^(1/3) −((2x−3))^(1/3) = 0 x=? ((x−1))^(1/3) =a, ((x−2))^(1/3) =b, ⇒a^3 +b^3 =2x−3 ⇒a+b=((a^3 +b^3 ))^(1/3) ⇒(a+b)^3 =a^3 +b^3 ⇒3ab(a+b)=0 ⇒ab(a+b)=0 a=0 ⇒x=1 b=0 ⇒x=2 a+b=0 ⇒((2x−3))^(1/3) =0 ⇒x=(3/2)
$$\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}}\:−\sqrt[{\mathrm{3}}]{\mathrm{2x}−\mathrm{3}}\:=\:\mathrm{0} \\ $$$$\:\:\:\mathrm{x}=? \\ $$$$\:\:\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:={a},\:\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}={b}, \\ $$$$\:\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{2}{x}−\mathrm{3} \\ $$$$\:\Rightarrow{a}+{b}=\sqrt[{\mathrm{3}}]{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} } \\ $$$$\:\Rightarrow\left({a}+{b}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$$$\:\Rightarrow\mathrm{3}{ab}\left({a}+{b}\right)=\mathrm{0}\:\Rightarrow{ab}\left({a}+{b}\right)=\mathrm{0} \\ $$$$\:{a}=\mathrm{0}\:\Rightarrow{x}=\mathrm{1} \\ $$$$\:{b}=\mathrm{0}\:\Rightarrow{x}=\mathrm{2} \\ $$$$\:{a}+{b}=\mathrm{0}\:\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{3}}=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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