các bạn giúp mik với!! (làm sao cho thông minh chứ đừng khai triển ra nhé)
$(x +1)^3+ (2x +1)^3+ 3(2x2 +3x +1)(6x+5) = (3x+2)^3$
CM $ a^3 + b^3 + c^3 = 3abc $ khi $ a + b + c = 0 $
$ (x + 1)^3 + (2x + 1)^3 + 3(2x^2 + 3x + 1)(6x + 5) = (3x + 2)^3 \\ \Leftrightarrow (x + 1)^3 + (2x + 1)^3 - (3x + 2)^3 = - 3(2x^2 + 3x + 1)(6x + 5)\\ \Leftrightarrow \\ (x + 1)^3 + (2x + 1)^3 + [-(3x + 2)]^3 = - 3(2x^2 + 3x + 1)(6x + 5)\\ (x + 1) + (2x + 1) + [-(3x + 2)] = 0 \\ \Rightarrow (x + 1)^3 + (2x + 1)^3 + [-(3x + 2)]^3 = -3(x + 1)(2x + 1)(3x + 2) = - 3(2x^2 + 3x + 1)(6x + 5) \\ \Rightarrow -3(x + 1)(2x + 1)(3x + 2) = -3(x + 1)(2x + 1)(6x + 5) \\\Leftrightarrow 3(x + 1)(2x + 1)(-3x - 2 + 6x + 5) = 0 \\ \Leftrightarrow -3(x + 1)(2x + 1)(3x + 3) = 0 \\ \Leftrightarrow \left[\begin{matrix}
x + 1 = 0\\
2x + 1 = 0\\
3x + 3 = 0
\end{matrix}\right. \\\Leftrightarrow \left[\begin{matrix}
x = -1\\
x = \dfrac{-1}{2}
\end{matrix}\right. $
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