Ta có (a3 + b3) + c3 - 3abc = 0
<=> (a + b)3 - 3ab(a + b) + c3 - 3abc = 0
<=> (a + b + c)[(a + b)2 - (a + b)c + c2] - 3ab(a + b + c) = 0
<=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
<=> (a + b + c).(2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc) = 0
<=> (a + b + c)[(a - b)2 + (b - c)2 + (c - a)2] = 0 (1)
Áp dụng (1) cho bài toán ta được
(x - 1)3 + (2x - 3)3 + (3x - 5)3 - 3(x - 1)(2x - 3)(3x - 5) = 0
<=> (6x - 9)[(x - 2)2 + (x - 2)2 + (2x - 4)2] = 0
<=> \(\left[{}\begin{matrix}6x-9=0\\\left(x-2\right)^2+\left(x-2\right)^2+\left(2x-4\right)^2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\6.\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
<=> (a + b)3 - 3ab(a + b) + c3 - 3abc = 0
<=> (a + b + c)[(a + b)2 - (a + b)c + c2] - 3ab(a + b + c) = 0
<=> (a + b + c)(a2 + b2 + c2 - ab - ac - bc) = 0
<=> (a + b + c).(2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc) = 0
<=> (a + b + c)[(a - b)2 + (b - c)2 + (c - a)2] = 0 (1)
Áp dụng (1) cho bài toán ta được
(x - 1)3 + (2x - 3)3 + (3x - 5)3 - 3(x - 1)(2x - 3)(3x - 5) = 0
<=> (6x - 9)[(x - 2)2 + (x - 2)2 + (2x - 4)2] = 0
<=> 
<=>
