Bonjour Ouiamita12341,
Puisque les racines cubiques n'imposent aucune condition, le domaine de chacune des équations est l'ensemble R des réels.
[tex]1)\ \sqrt[3]{3x+1}-1=x\\\\\sqrt[3]{3x+1}=x+1\\\\(\sqrt[3]{3x+1})^3=(x+1)^3\\\\3x+1=(x+1)^3\\\\3x+1=x^3+3x^2+3x+1\\\\x^3+3x^2+3x+1-3x-1=0\\\\x^3+3x^2=0\\\\x^2(x+3)=0\\\\x^2=0\ \ ou\ \ x+3=0\\\\\boxed{x=0\ \ ou\ \ x=-3}[/tex]
[tex]2)\ \sqrt[3]{1-6x}+x=1\\\\\sqrt[3]{1-6x}=1-x\\\\(\sqrt[3]{1-6x})^3=(1-x)^3\\\\1-6x=(1-x)^3\\\\1-6x=1-3x+3x^2-x^3\\\\x^3-3x^2-3x=0\\\\x(x^2-3x-3)=0\\\\x=0\ \ ou\ \ x^2-3x-3=0\\\\a)\ \boxed{x=0}\\\\b)\ x^2-3x-3=0\\\Delta=(-3)^2-4\times1\times(-3)=9+12=21\ \textgreater \ 0\\\\\boxed{x_1=\dfrac{3-\sqrt{21}}{2}}\ \ ou\ \ \boxed{x_2=\dfrac{3+\sqrt{21}}{2}}\\\\\\\Longrightarrow\ \text{Les solutions sont donc : }\\\\\boxed{x=0\ \ ou\ \ x=\dfrac{3-\sqrt{21}}{2}\ \ ou\ \ x=\dfrac{3+\sqrt{21}}{2}}[/tex]
[tex]3)\ \sqrt[3]{2x-x^2}=\dfrac{1}{2}\\\\(\sqrt[3]{2x-x^2})^3=(\dfrac{1}{2})^3\\\\2x-x^2=\dfrac{1}{8}\\\\8\times(2x-x^2)=8\times(\dfrac{1}{8})\\\\16x-8x^2=1\\\\8x^2-16x+1=0\\\\\Delta=(-16)^2-4\times8\times1=256-32=224\ \textgreater \ 0\\\\x_1=\dfrac{16-\sqrt{224}}{2\times8}=\dfrac{16-4\sqrt{14}}{16}=\dfrac{4(4-\sqrt{14})}{16}=\dfrac{4-\sqrt{14}}{4}\\\\x_2=\dfrac{16+\sqrt{224}}{2\times8}=\dfrac{16+4\sqrt{14}}{16}=\dfrac{4(4+\sqrt{14})}{16}=\dfrac{4+\sqrt{14}}{4}\\\\\\\Longrightarrow\ \text{Les solutions sont donc : }[/tex]
[tex]\boxed{x=\dfrac{4-\sqrt{14}}{4}\ \ ou\ \ x=\dfrac{4+\sqrt{14}}{4}}[/tex]
[tex]4)\ 3\sqrt[3]{1-x}+x-1=0\\\\3\sqrt[3]{1-x}=1-x\\\\\sqrt[3]{3^3}\sqrt[3]{1-x}=\sqrt[3]{(1-x)^3}\\\\\sqrt[3]{27(1-x)}=\sqrt[3]{(1-x)^3}\\\\27(1-x)=(1-x)^3\\\\(1-x)^3-27(1-x)=0\\\\(1-x)[(1-x)^2-27]=0\\\\1-x=0\ \ ou\ (1-x)^2-27=0\\\\x=1\ \ ou\ (1-x)^2=27\\\\x=1\ \ ou\ 1-x=-\sqrt{27}\ \ ou\ 1-x=\sqrt{27}\\\\x=1\ \ ou\ -x=-1-\sqrt{27}\ \ ou\ -x=-1+\sqrt{27}\\\\x=1\ \ ou\ x=1+\sqrt{27}\ \ ou\ x=1-\sqrt{27}\\\\\boxed{x=1\ \ ou\ x=1+3\sqrt{3}\ \ ou\ x=1-3\sqrt{3}}[/tex]
[tex]5)\ \sqrt[3]{x^2+8}=x+2\\\\(\sqrt[3]{x^2+8})^3=(x+2)^3\\\\x^2+8=x^3+6x^2+12x+8\\\\x^3+5x^2+12x=0\\\\x(x^2+5x+12)=0\\\\x=0\ \ ou\ \ x^2+5x+12=0\\\\x=0\ \ ou\ \ \Delta=5^2-4\times1\times12=25-48=-23\ \textless \ 0\ (pas\ de\ racine)\\\\\\\Longrightarrow\boxed{x=0}[/tex]
