Ta có: a+b+c=0 \(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)

\(\Rightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)

\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)(*)

Thay \(a+b=-c\) vào (*), có:

\(a^3+b^3+c^3=3abc\left(đpcm\right)\)

=1²-2.1.5a+ (5a)²

=1-10a+25a²

• Bài 1. 

a) \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow\left(25x^2+10x+1\right)-25x^2+9=30\)

\(\Leftrightarrow10x=20\Leftrightarrow x=2\)

b) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x-5=0\)

\(\Leftrightarrow4x=6\Leftrightarrow x=\frac{3}{2}\)

• Ta có : \(A=1997.1999=\left(1998-1\right)\left(1998+1\right)=1998^2-1< 1998^2\)

\(\Rightarrow A< B\)

• Từ a+b+c=2p => \(p=\frac{a+b+c}{2}\)

Ta có : \(4p\left(p-a\right)=2\left(a+b+c\right)\left(\frac{a+b+c}{2}-a\right)=2.\left(a+b+c\right).\frac{b+c-a}{2}\)

\(=\left(a+b+c\right)\left(b+c-a\right)=\left[\left(b+c\right)+a\right]\left[\left(b+c\right)-a\right]=\left(b+c\right)^2-a^2\)

\(=b^2+c^2-a^2+2bc\)

Bài cuối bạn sửa 2ab thành 2bc nhé ^^