\(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)
\(\Leftrightarrow\sqrt[3]{x+3}-2-\left(\sqrt[3]{6-x}-1\right)=0\)
\(\Leftrightarrow\dfrac{x+3-8}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}-\dfrac{6-x-1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}=0\)
\(\Leftrightarrow\dfrac{x-5}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{x-5}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}\right)=0\)
Dễ thấy: \(\dfrac{1}{\sqrt[3]{x+3}^2+4+2\sqrt[3]{x+3}}+\dfrac{1}{\sqrt[3]{6-x}^2+1+\sqrt[3]{6-x}}>0\)
\(\Rightarrow x-5=0\Leftrightarrow x=5\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+3}=a\\\sqrt[3]{6-x}=b\end{matrix}\right.\)thì co hệ
\(\left\{{}\begin{matrix}a=1+b\left(1\right)\\a^3+b^3=9\left(2\right)\end{matrix}\right.\)
\(\Rightarrow\left(1+b\right)^3+b^3=9\)
\(\Leftrightarrow\left(b-1\right)\left(2b^2+5b+8\right)=0\)
Dễ thây \(2b^2+5b+8>0\)
\(\Rightarrow b=1\)
\(\Rightarrow\sqrt[3]{6-x}=1\)
\(\Leftrightarrow x=5\)
\(pt\Leftrightarrow\sqrt[3]{x+3}=\sqrt[3]{6-x}+1\)
\(\Leftrightarrow2x-4=3\sqrt[3]{6-x}\left(\sqrt[3]{6-x}+1\right)\)
\(\Leftrightarrow2x-4=3\sqrt[3]{6-x}\sqrt[3]{x+3}\)
\(\Leftrightarrow8x^3-32x^2+64x-64=27\left(6-x\right)\left(x+3\right)\)
\(\Rightarrow...\)
\(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)
\(\Leftrightarrow\sqrt[3]{x+3}=\sqrt[3]{6-x}+1\)
\(\Leftrightarrow2x-4=3.\sqrt[3]{6-x}\left(\sqrt[3]{6-x}+1\right)\)
\(\Leftrightarrow2x-4=3.\sqrt[3]{6-x}.\sqrt[3]{x+3}\)
\(\Leftrightarrow8x^3-32x^2+64x-64=27\left(6-x\right)\left(x+3\right)\)
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