Ta có: a³+b³+c³=3abc <=> a³+b³+c³-3abc=0

<=>\(a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)

<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

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

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

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

Mà a+b+c khác 0

=>\(a^2+b^2+c^2-ab-bc-ca=0\)

<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)

=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

- Ta có : \(a^3+b^3+c^3=3abc\)

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

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

Mà \(a+b+c\ne0\)

=> \(a^2+b^2+c^2-ab-bc-ac=0\)

=> \(\frac{\left(a^2-2ab+b^2\right)+\left(b^2-2ac+c^2\right)+\left(c^2-2ac+a^2\right)}{2}=0\)

=> \(\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=0\)

=> \(a-b=b-c=c-a=0\)

=> \(a=b=c\)

- Thay a = b = c vào biểu thức N ta được :

\(N=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy giá trị của N = \(\frac{1}{3}\) khi \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\)

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

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

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

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

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

Ta có : \(\frac{a+b-c}{c}=\frac{a+c-b}{b}=\frac{b+c-a}{a}\)

\(\Rightarrow\frac{a+b}{c}-\frac{c}{c}=\frac{a+c}{b}-\frac{b}{b}=\frac{b+c}{a}-\frac{a}{a}\)

\(\frac{a+b}{c}-1=\frac{c+b}{a}-1=\frac{a+c}{b}-1\)

\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có

       \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)

Vậy \(P=\left(a+b\right)\left(b+c\right)\left(c+a\right)=2c.2a.2b=8abc\)

mà \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=abc\Rightarrow8abc=abc\Rightarrow abc=0\Rightarrow P=0\)

Áp dụng bđt AM - GM:

\(T=\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}=\left(\dfrac{1}{9}\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\right)+\dfrac{8}{9}\dfrac{a+b+c}{\sqrt[3]{abc}}\ge2\sqrt{\dfrac{1}{9}}+\dfrac{8}{9}.3=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\).

Đẳng thức xảy ra khi a = b = c.

Vậy Min T = \(\dfrac{10}{3}\) khi a = b = c.

(a+b-c)/c+2 =(b+c-a)/c+2 =(c+a-b)/c+2 

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