Ta có : \(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)+c^3=0\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab.-c=3abc\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

1. Cho \(a^3+b^3+c^3=3abc\Rightarrow a+b+c=0\)

Ta có : \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3+c^3-3ab\left(a+b\right)=3abc\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

Vậy \(a^3+b^3+c^3=3abc\Rightarrow a+b+c=0\)(1)

           2. Cho \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)

Xét \(a^3+b^3+c^3-3abc\)

\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3+c^3-3ab\left(a+b\right)-3abc\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

mà \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Vậy \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)(2)

Từ (1)(2)\(\Rightarrow a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\)(ĐPCM)

Đặt  \(n^4+n^3+n^2=a^2\left(a\in N\right)\)

Ta có : \(n^4-2n^3+n^2< a^2< n^4+2n^3+n^2\) 

\(\Leftrightarrow\left(n^2-n\right)^2< a^2< \left(n^2+n\right)^2\)\(\Rightarrow n^2-n< a< n^2+n\)

Mặt khác, ta lại có : \(n^2-n< n^2< n^2+n\) \(\Rightarrow a=n^2\Leftrightarrow a^2=n^4\)

\(\Leftrightarrow n^4+n^3+n^2=n^4\Leftrightarrow n^2\left(n+1\right)=0\Leftrightarrow\orbr{\begin{cases}n=0\left(\text{nhận}\right)\\n=-1\left(\text{loại}\right)\end{cases}}\)

Vậy n = 0 thoả mãn đề bài.

ko co them dieu kien a ban

a ) ( x³ -x + 3x³y + 3xy² + y³ -y) = ( x + y )³ - ( x + y ) = ( x-y )² ( x - y - 1 )

b) x² + 5x -6 = x² + 6x -x - 6 = x( x +  6 ) - ( x + 6 ) = ( x -1 ) ( x + 6 )

c) 16 x - 5x² - 3 = -5x² + 15x +x -3 = -5x ( x-3 ) + ( x - 3 ) = ( 1 - 5x ) ( x-3)