Sai đề! Sửa: that 2c+b-a=2c+a-b

Đặt 2a+b-c=x, 2b+c-a=y, 2c+a-b=z

\(\Rightarrow8\left(a+b+c\right)^3=\left(x+y+z\right)^3=x^3+y^3+z^3\)và \(P=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có: \(\left(x+y+z\right)^3-x^3-y^3-z^3=0\Leftrightarrow\left(x+y\right)^3+3\left(x+y\right)z\left(x+y+z\right)-x^3-y^3=0\)

\(\Leftrightarrow3xy\left(x+y\right)+3\left(x+y\right)z\left(x+y+z\right)=0\Leftrightarrow3\left(x+y\right)\left(xy+xz+yz+z^2\right)=0\)

\(\Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\Leftrightarrow3P=0\Leftrightarrow P=0\)

a) \(A+B=-12x^2y^4-6x^2y^4=-18x^2y^4\)

\(A+C=-12x^2y^4+9x^2y^4=-3x^2y^4\)

\(B+C=-6x^2y^4+9x^2y^4=3x^2y^4\)

a) 

\(A=\left(\left(a+b\right)+c\right)^3+\left(\left(a-b\right)-c\right)^3-ba\left(b+c\right)^2\\ \)

\(A=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)\(-\left(a-b\right)^3-3\left(a-b\right)^2c+3\left(a-b\right)c^2-c^3\)\(-ba\left(b^2+2bc+c^2\right)\)

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

\(A=a^3+3a^2b+3ab^2+b^3+3a^2c+6abc+3b^2c\)\(+3ac^2+3bc^2+c^3-a^3+3a^2b-3ab^2+b^3\)\(-3a^2c+6abc-ab^2c+3ac^2-3bc^2-c^3-b^3a\)\(-2ab^2c-bac^2\)

\(A=6a^2b+2b^3+12abc+3b^2c-3ab^2c-c^3-b^3a-bac^2\)

ko bt có đúng ko nữa

nếu sai cho mình xin lỗi nha

\(a^3+b^3+c^3-3abc=1\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=1\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=1\) (1)

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

(1)\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca+\dfrac{1}{a+b+c}\)

\(\Leftrightarrow3a^2+3b^2+3c^2=\left(a+b+c\right)^2+\dfrac{1}{a+b+c}\ge3\)

\(\Rightarrow a^2+b^2+c^2\ge1\)

Sai đề!