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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)

*) Xét a + b + c = 0 => \(\hept{\begin{cases}-a=b+c\\-b=c+a\\-c=a+b\end{cases}}\)

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)

*) Xét \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Vậy A = -1 hoặc A = 8

a. Câu hỏi của Nguyễn Thị Anh Thư - Toán lớp 8 - Học toán với OnlineMath

a, \(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)

\(=-2\left[x^2-2x+1+x^2-1+x^2+2x+1\right]+6x^2-6\)

\(=-2\left(3x^2+1\right)+6x^2-6=-6x^2-2+6x^2-6=-8\)

b, \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)

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

\(=\left(x-1\right)\left[\left(x-1\right)^2-\left(x^2+x+1\right)+3x\right]\)

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

\(=\left(x-1\right).0=0\)

a ) 

\(A=xy\left(3x^2-6xy\right)-3\left(x^3y-2x^2y^2-1\right)\)

\(\Leftrightarrow A=3x^3y-6x^2y^2-3x^3y+6x^2y^2+3\)

\(\Leftrightarrow A=3\)

\(\Leftrightarrow A\)ko phụ thuộc vào g/t của biến 

b ) 

\(B=\left(x-9\right)\left(x-9\right)+\left(2x+1\right)^2-\left(5x-4\right)\left(x-2\right)\)

\(\Leftrightarrow B=x^2-2.x.9+9^2+\left(2x\right)^2+2.2x.1+1-\left[5x^2-4x-10x+8\right]\)

\(\Leftrightarrow B=x^2-18x+81+4x^2+4x+1-5x^2+4x+10x-8\)

\(\Leftrightarrow B=\left(x^2+4x^2-5x^2\right)+\left(-18x+4x+4x+10x\right)+\left(81-8+1\right)\)

\(\Leftrightarrow B=74\)

\(\Leftrightarrow B\)ko phụ thuộc vào g/t của biến 

\(a+b+c=0\)

\(a^3+b^3+c^3=3.\overline{abc}\)

_______________________

Ta xét vế trái \(a^3+b^3+c^3=\left[\left(a+b\right)\left(a^2-ab+b^2\right)\right]+c^3\)\(\left(1\right)\)
Mà  theo giả thiết ta có : \(a+b+c=0\) \(\Rightarrow c=a+b\Rightarrow c^3=-\left(a+b\right)^3\)
Thay vào \(\left(1\right)\)Ta có :\(\text{ [(a+b)(a^2-ab+b^2)] - (a+b)^3}\)
                                      \(\text{=(a+b)[a^2-ab+b^2-(a+b)^2] }\)( lấy nhân tử chung)
\(\text{ =(a+b)^2) =(a+b)[a^2-ab+b^2-(a^2+2ab+b^2)] }\)( phân tích )
\(\text{=(a+b)(a^2-ab+b^2-a^2-2ab-b^2) }\)
\(\text{=(a+b).(-3ab) }\)
\(\text{= -(a+b).3ab (2) }\)
Theo giả thiết ta có :\(\text{ a+b+c=0 \Rightarrow c= -(a+b) }\)
Thay vào \(\left(2\right)\)ta được \(=3abc\)(đpcm)

 

Sau a^3 là dấu "+" đúng ko ?

Ta có: \(a+b+c=0\Leftrightarrow c=-a-b\)

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

\(=-3a^2b-3ab^2=3ab\left(-a-b\right)\)

Lại có: \(-a-b=c\Rightarrow a^3+b^3+c^3=3abc\)(đpcm).