1.Tìm x:
\(\left(x-1\right)^3+\left(2x+1\right)^3+\left(x+2\right)^3=3\left(x-1\right).\left(2x+1\right).\left(x+2\right)\)
2. Cho \(a+b+c=0\) . C/m: \(a^3+b^3+c^3=3abc\)
3. Tìm x:
a, \(\left(x+1\right)^3+\left(2x-3\right)^3+\left(2-3x\right)^3=0\)
b, \(\left(2x+1\right)^3+\left(x+2\right)^3=27\left(x+1\right)^3\)
2. \(a+b+c=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3a^{2c}+3ac^2+3b^2c+3bc^2+6abc\)
\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)-3abc\)
\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3ac\left(a+c+b\right)+3bc\left(b+c+a\right)-3abc\)
Ta có: \(a+b+c=0\)
\(a^3+b^3+c^3+3ab.0+3ac.0+3bc.0=3abc\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
Bài 2
\(a+b+c=0\Rightarrow a=-b-c\)
\(VT=a^3+b^3+c^3=\left(-b-c\right)^3+b^3+c^3\)
\(=\left(-b\right)^3-3\left(-b\right)^2c+3\left(-b\right)c^2-c^3+b^3+c^3\)
\(=\left(-b\right)^3-3b^2c-3bc^2-c^3+b^3+c^3\)
\(=-3b^2c-3bc^2=3bc\left(-b-c\right)=3abc=VP\)
bài 2
ta có a+b+c=0
=>a+b=-c
=>c=-(a+b)
thay -(a+b)=c vào 2 vế ta đc
a3+b3-(a+b)3=3ab[-a-b)]
=>a3+b3-(a3+3a2b+3ab2+b3)=-3a2b-3ab2
=>a3+b3-a3-3a2b-3ab2-b2=-3ab(a-b)
=>(a3-a3)+(b3-b3)+(-3a2b-3ab2)=-3ab(a-b)
=>0+0-3ab(a-b)=-3ab(a-b)(đpcm)