\(\sqrt[3]{x+1}=x^3-15x^2+75x-125-6=0\)
\(\Leftrightarrow\sqrt[3]{x+1}+6=\left(x-5\right)^3\)
Đặt \(\sqrt[3]{x+1}=a-5\) ta được hệ:
\(\left\{{}\begin{matrix}\left(a-5\right)^3=x+1\\a-5+6=\left(x-5\right)^3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-5\right)^3=x+1\\\left(x-5\right)^3=a+1\end{matrix}\right.\)
Trừ vế cho vế ta được:
\(\left(x-5\right)^3-\left(a-5\right)^3=a-x\)
\(\Leftrightarrow\left(x-a\right)\left(\left(x-5\right)^2+\left(x-5\right)\left(a-5\right)+\left(a-5\right)^2\right)+\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left[\left(x-5+\frac{a-5}{2}\right)^2+\frac{3\left(a-5\right)^2}{4}+1\right]=0\)
\(\Leftrightarrow x-a=0\) (phần ngoạc phía sau luôn dương)
\(\Leftrightarrow x=a\Leftrightarrow x=\sqrt[3]{x+1}+5\Leftrightarrow x-5=\sqrt[3]{x+1}\)
\(\Leftrightarrow x^3-15x^2+75x-125=x+1\)
\(\Leftrightarrow x^3-15x^2+74x-126=0\)
\(\Rightarrow x=7\)
