Ta chứng minh tính chất sau: với các số thực \(a;b;c\) sao cho \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)
Thật vậy ta có: \(a+b+c=0\Rightarrow a+b=-c\)
\(a^3+b^3+c^3+3ab\left(a+b\right)-3ab\left(a+b\right)=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left(\left(a+b^2\right)-\left(a+b\right)c+c^2\right)-3ab\left(-c\right)\)
\(=-3ab\left(-c\right)=3abc\) (đpcm)
Áp dụng cho bài toán:
\(\left(x^2-3x+2\right)^3-x^6+\left(3x-2\right)^3=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)^3+\left(-x^2\right)^3+\left(3x-2\right)^3=0\) (1)
Do \(x^2-3x+2+\left(-x^2\right)+3x-2=0\)
\(\Rightarrow\left(x^2-3x+2\right)^3+\left(-x^2\right)^3+\left(3x-2\right)^3=3\left(x^2-3x+2\right)\left(-x^2\right)\left(3x-2\right)\)
Phương trình (1) trở thành:
\(\left(x^2-3x+2\right)\left(-x^2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\\-x^2=0\\3x-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\\x=\frac{2}{3}\end{matrix}\right.\)
\(x^6-\left(3x-2\right)^3=\left(x^2-3x+2\right)\left[x^4+x^2\left(3x-2\right)+\left(3x-2\right)^2\right]\)
Nhân tử chung 2 vế: x^2-3x+2. Giải pt đó nha
\(\left(x^2-3x+2\right)^3=x^6-\left(3x-2\right)^3\\ \Leftrightarrow\left(x^2-x-2x+2\right)^3=\left[x^2-\left(3x-2\right)\right]\left[x^4+x^2\left(3x-2\right)+\left(3x-2\right)^2\right]\\ \Leftrightarrow\left[x\left(x-1\right)-2\left(x-1\right)\right]^3=\left(x^2-3x+2\right)\left(x^4+3x^3-2x^2+9x^2-12x+4\right)\\ \Leftrightarrow\left[\left(x-1\right)\left(x-2\right)\right]^3=\left(x^2-x-2x+2\right)\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)^3\left(x-2\right)^3=\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)^3\left(x-2\right)^3=\left(x-1\right)\left(x-2\right)\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)^3\left(x-2\right)^3-\left(x-1\right)\left(x-2\right)\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left\{\left[\left(x-1\right)\left(x-2\right)\right]^2-x^4-3x^3-7x^2+12x-4\right\}=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left[\left(x^2-2x-x+2\right)^2-x^4-3x^3-7x^2+12x-4\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left[\left(x^2-3x+2\right)^2-x^4-3x^3-7x^2+12x-4\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\\left(x^2-3x+2\right)^2-x^4-3x^3-7x^2+12x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\\x=\frac{2}{3}\end{matrix}\right.\)
