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\(x^3=2a+3x.\sqrt[3]{a^2-\frac{\left(a+1\right)^2}{9}\left(\frac{8a-1}{3}\right)}\)
\(x^3=2a+3x\sqrt[3]{\frac{1-6a+12a^2-8a^3}{27}}\)
\(x^3=2a+3x\sqrt[3]{\left(\frac{1-2a}{3}\right)^3}\)
\(x^3=2a+\left(1-2a\right)x\)
\(x^3-x+2ax-2a=0\)
\(x\left(x^2-1\right)+2a\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^2+x+2a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+x+2a=0\left(1\right)\end{matrix}\right.\)
Xét (1): \(x^2+x+\frac{1}{4}+2a-\frac{1}{4}=0\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)=0\)
Do \(a>\frac{1}{8}\Rightarrow\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)>0\)
\(\Rightarrow\left(1\right)\) vô nghiệm \(\Rightarrow x=1\) hay x nguyên dương
\(x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\)
\(\Leftrightarrow x^3=\left(\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)^3\)
\(\Leftrightarrow x^3=2a+3.\sqrt[3]{a^2-\frac{\left(a+1\right)^2}{9}.\frac{8a-1}{3}}.\left(\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\right)\)
\(\Leftrightarrow x^3=2a+3.\sqrt[3]{\frac{-8a^3+12a^2+6a-1}{27}}.x\)
\(\Leftrightarrow x^3=2a+3.\sqrt[3]{-\left(\frac{2a-1}{3}\right)^3}.x\)
\(\Leftrightarrow x^3=2a-\left(2a-1\right)x\Leftrightarrow x^3+\left(2a-1\right)x-2a=0\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)+\left(2a-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+2a\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x^2+x+2a=0\end{array}\right.\) . Ta có : \(x^2+x+2a=\left(x^2+x+\frac{1}{4}\right)+2\left(a-\frac{1}{8}\right)=\left(x+\frac{1}{2}\right)^2+2\left(a-\frac{1}{8}\right)\ge2\left(a-\frac{1}{8}\right)\)
Vì \(a>\frac{1}{8}\Rightarrow x^2+x+2a>0\) => vô nghiệm.
Vậy x = 1 => x là số tự nhiên.
\(x^3=2a+3x\sqrt[3]{a^2-\left(\frac{a+1}{3}\right)^2\left(\frac{8a-1}{3}\right)}\)
\(\Leftrightarrow x^3=2a+3x\cdot\frac{\sqrt[3]{\left(1-2a\right)^3}}{3}\)
\(\Leftrightarrow x^3=2a+x\left(1-2a\right)\)
\(\Leftrightarrow x^3+\left(2a-1\right)x-2a=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+2a\right)=0\)
\(\Leftrightarrow x-1=0\)(do \(x^2+x+2a\)vô nghiệm vì \(a>\frac{1}{8}\))
<=> x=1 nên là 1 số nguyên dương
Dấu ở giữa là cộng chứ nhỉ??
Đặt \(y=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}};z=\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}\)
\(\Rightarrow\hept{\begin{cases}y^3+z^3=2a\\yz=\sqrt[3]{a^2-\frac{\left(a+1\right)^2\left(8a-1\right)}{27}}\\y+z=x\end{cases}=\sqrt[3]{\frac{27a^2-\left(8a^3+15a^2+6a-1\right)}{27}}=\sqrt[3]{\frac{\left(1-2a\right)^3}{27}}=\frac{1-2a}{3}}\)
Thay vào ta được:
\(x^3=\left(y+z\right)^3=y^3+z^3+3yz\left(y+z\right)\)\(=2a+3\frac{1-2a}{3}x=2a+\left(1-2a\right)x\)
\(\Leftrightarrow x^3-\left(1-2a\right)x-2a=0\)
\(\Leftrightarrow x^3-x+2ax-2a=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2a+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+2a+x=0\end{cases}}\)
Đến đây thì có lẽ là sẽ cm được \(x^2+2a+x>0\), mình chưa tìm ra cách cm.
KL : \(x=1\inℤ\)