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Question Number 119848 by benjo_mathlover last updated on 27/Oct/20
Solve in real numbers the equation  (x)^(1/(3 ))  + ((x−1))^(1/(3 ))  + ((x+1))^(1/(3 ))  = 0
$${Solve}\:{in}\:{real}\:{numbers}\:{the}\:{equation} \\ $$$$\sqrt[{\mathrm{3}\:}]{{x}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}+\mathrm{1}}\:=\:\mathrm{0} \\ $$
Answered by 1549442205PVT last updated on 27/Oct/20
Applying the identity   a^3 +b^3 +c^3 =3abc when a+b+c=0.Then  (x)^(1/(3 ))  + ((x−1))^(1/(3 ))  + ((x+1))^(1/(3 ))  = 0 gives us  3x=3^3 (√(x(x^2 −1))) ⇔x^3 =x(x^2 −1)  ⇔x=0.Thus x=0 is unique root of  given equation
$$\mathrm{Applying}\:\mathrm{the}\:\mathrm{identity}\: \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} =\mathrm{3abc}\:\mathrm{when}\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{0}.\mathrm{Then} \\ $$$$\sqrt[{\mathrm{3}\:}]{{x}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}+\mathrm{1}}\:=\:\mathrm{0}\:\mathrm{gives}\:\mathrm{us} \\ $$$$\mathrm{3x}=\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}\:\Leftrightarrow\mathrm{x}^{\mathrm{3}} =\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\Leftrightarrow\mathrm{x}=\mathrm{0}.\mathrm{Thus}\:\mathrm{x}=\mathrm{0}\:\mathrm{is}\:\mathrm{unique}\:\mathrm{root}\:\mathrm{of} \\ $$$$\mathrm{given}\:\mathrm{equation} \\ $$$$ \\ $$

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