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Question Number 201966 by mnjuly1970 last updated on 17/Dec/23
         solve           ((x^2 +3x+2))^(1/3)  (((x+1))^(1/3)  −((x+2))^(1/3)  )= 1
$$ \\ $$$$\:\:\:\:\:\:\:{solve}\: \\ $$$$\:\: \\ $$$$\:\:\:\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}\:\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:\right)=\:\mathrm{1} \\ $$$$ \\ $$
Answered by ajfour last updated on 17/Dec/23
let  x+(3/2)=t  ⇒  p=x+1=t−(1/2)  q=x+2=t+(1/2)  ⇒ −R= x^2 +3x+2=t^2 −(1/4)  pq=−R  (t^2 −(1/4))[−1−((3(pq)^(1/3) )/((t^2 −(1/4))^(1/3) ))]=1  −1−((3(pq)^(1/3) )/((t^2 −(1/4))^(1/3) ))=  −(1/R)  ((1/R)−1)^3 =((27(−R))/(−R))  ⇒  (1/R)−1=3  ⇒  R=(1/4)       x^2 +3x+2=−(1/4)  x^2 +3x+(9/4)=0  (x+(3/2))^2 =0  x= −(3/2)
$${let}\:\:{x}+\frac{\mathrm{3}}{\mathrm{2}}={t}\:\:\Rightarrow\:\:{p}={x}+\mathrm{1}={t}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${q}={x}+\mathrm{2}={t}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:−{R}=\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}={t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${pq}=−{R} \\ $$$$\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)\left[−\mathrm{1}−\frac{\mathrm{3}\left({pq}\right)^{\mathrm{1}/\mathrm{3}} }{\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} }\right]=\mathrm{1} \\ $$$$−\mathrm{1}−\frac{\mathrm{3}\left({pq}\right)^{\mathrm{1}/\mathrm{3}} }{\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} }=\:\:−\frac{\mathrm{1}}{{R}} \\ $$$$\left(\frac{\mathrm{1}}{{R}}−\mathrm{1}\right)^{\mathrm{3}} =\frac{\mathrm{27}\left(−{R}\right)}{−{R}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{{R}}−\mathrm{1}=\mathrm{3}\:\:\Rightarrow\:\:{R}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 17/Dec/23
  thanks alot sir ajfor
$$\:\:{thanks}\:{alot}\:{sir}\:{ajfor} \\ $$$$\:\: \\ $$
Answered by Frix last updated on 17/Dec/23
(ab)^(1/3) (a^(1/3) −b^(1/3) )=1  a^(1/3) −b^(1/3) _(★) =(ab)^(−(1/3))   a−3(ab)^(1/3) (a^(1/3) −b^(1/3) _(★) )−b=(ab)^(−1)   a−b−3=(1/(ab))  a=x+1∧b=x+2  x^2 +3x+(9/4)=0  x=−(3/2)  [Only true if we stay ∈R ⇒ ((−r))^(1/3) =−(r)^(1/3) ]
$$\left({ab}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)=\mathrm{1} \\ $$$$\underset{\bigstar} {\underbrace{{a}^{\frac{\mathrm{1}}{\mathrm{3}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} }}=\left({ab}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${a}−\mathrm{3}\left({ab}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\underset{\bigstar} {\underbrace{{a}^{\frac{\mathrm{1}}{\mathrm{3}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} }}\right)−{b}=\left({ab}\right)^{−\mathrm{1}} \\ $$$${a}−{b}−\mathrm{3}=\frac{\mathrm{1}}{{ab}} \\ $$$${a}={x}+\mathrm{1}\wedge{b}={x}+\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left[\mathrm{Only}\:\mathrm{true}\:\mathrm{if}\:\mathrm{we}\:\mathrm{stay}\:\in\mathbb{R}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{−{r}}=−\sqrt[{\mathrm{3}}]{{r}}\right] \\ $$
Commented by Frix last updated on 17/Dec/23
...and no complex solutions.
$$…\mathrm{and}\:\mathrm{no}\:\mathrm{complex}\:\mathrm{solutions}. \\ $$

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