\(\left(x-1\right)^3+\left(x-3\right)^3=\left(2x-4\right)^3\)
\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3-\left(2x-4\right)^3=0\)
\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3+\left(4-2x\right)^3=0\)
Đặt \(\left(x-1\right)=a;\left(x-3\right)=b;\left(4-2x\right)=c\)ta có:
\(a^3+b^3+c^3=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Thay lại, ta được:
\(\left(a+b+c\right)^3=\left(x-1+x-3+4-2x\right)^3=0\)
\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(x-1+x-3\right)\left(x-3+4-2x\right)\left(4-2x+x-1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(1-x\right)\left(3-x\right)=0\)
\(\Leftrightarrow2x-4=0\Leftrightarrow2x=4\Rightarrow x=2\)
hoặc \(1-x=0\Leftrightarrow x=1\)
hay \(3-x=0\Leftrightarrow x=3\)
Vậy \(x\in\left\{1;2;3\right\}\). Xong :))
