Bài 1: 

$a^3+b^3+c^3=3abc$

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

$\Leftrightarrow [(a+b)^3+c^3]-[3ab(a+b)+3abc]=0$

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

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

$\Rightarrow a+b+c=0$ hoặc $a^2+b^2+c^2-ab-bc-ac=0$

Xét TH $a^2+b^2+c^2-ab-bc-ac=0$

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

$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\Rightarrow a-b=b-c=c-a=0$

$\Leftrightarrow a=b=c$

Vậy $a^3+b^3+c^3=3abc$ khi $a+b+c=0$ hoặc $a=b=c$

Áp dụng vào bài:

Nếu $a+b+c=0$

$A=\frac{-c}{c}+\frac{-b}{b}+\frac{-a}{a}=-1+(-1)+(-1)=-3$

Nếu $a=b=c$

$P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2+2+2=6$

3)

1.

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với \(\dfrac{2}{3}\), không mất tính tổng quát, giả sử đó là b và c

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\ge0\)

Mặt khác \(0\le a\le1\Rightarrow1-a\ge0\)

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\left(1-a\right)\ge0\)

\(\Leftrightarrow-abc\ge\dfrac{4a}{9}+\dfrac{2b}{3}+\dfrac{2c}{3}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}\)

\(\Leftrightarrow-abc\ge-\dfrac{2a}{9}+\dfrac{2}{3}\left(a+b+c\right)-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}=-\dfrac{2a}{9}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc+\dfrac{8}{9}\)

\(\Leftrightarrow-2abc\ge-\dfrac{4a}{9}-\dfrac{4ab}{3}-\dfrac{4ac}{3}-2bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{ab}{3}-\dfrac{ac}{3}-bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(b+c\right)-bc+\dfrac{16}{9}\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(2-a\right)-\dfrac{\left(b+c\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}+\dfrac{a^2}{3}-\dfrac{2a}{3}-\dfrac{\left(2-a\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge\dfrac{a^2}{12}-\dfrac{a}{9}+\dfrac{7}{9}=\dfrac{1}{12}\left(a-\dfrac{2}{3}\right)^2+\dfrac{20}{27}\ge\dfrac{20}{27}\)

\(\Rightarrow ab+bc+ca\ge2abc+\dfrac{20}{27}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)

2.

Đặt \(\left(a;b;c\right)=\left(x+1;y+1;z+1\right)\Rightarrow\left\{{}\begin{matrix}x;y;z\in\left[0;2\right]\\x+y+z=3\end{matrix}\right.\)

Ta có: \(P=\left(x+1\right)^3+\left(y+1\right)^3+\left(z+1\right)^3\)

\(P=x^3+y^3+z^3+3\left(x^2+y^2+z^2\right)+12\)

Không mất tính tổng quát, giả sử \(x\ge y\ge z\Rightarrow x\ge1\)

\(\Rightarrow\left\{{}\begin{matrix}y^3+z^3=\left(y+z\right)^3-3yz\left(y+z\right)\le\left(y+z\right)^3\\y^2+z^2=\left(y+z\right)^2-2yz\le\left(y+z\right)^2\end{matrix}\right.\)

\(\Rightarrow P\le x^3+\left(3-x\right)^3+3x^2+3\left(3-x\right)^2+12\)

\(\Rightarrow P\le15x^2-45x+66=15\left(x-1\right)\left(x-2\right)+36\le36\)

(Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\))

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;1;0\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(1;2;3\right)\) và các hoán vị

a(b³ - c³) + b(c³ - a³) + c(a³ - b³)

= a(b³ - c³ ) + b( c³ - b³ + b³ - a³) + c(a³ - b³)

= a(b³ - c³) + b(c³ - b³) + b(b³ - a³) + c(a³ - b³)

= a(b³ - c³) - b(b³ - c³) - [b(a³ - b³) - c(a³- b³)]

= (b³ - c³)(a - b) - (a³- b³)(b - c)

= (b - c)(b² + bc + c²)(a - b) - (a - b)(a² + ab + b²)(b - c)

= (b - c)(a - b)(b² + bc + c² - a² + ab - b²)

= (b - c)(a - b) [ (c²  - a²) + (bc - ab) ]

= (b - c)(a - b) [ (c - a)(c + a) + b(c - a) ]

= (b - c)(a -b) [ (c - a)(c + a + b) ]

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

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Trường hợp 1: a+b+c=0

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2=b^2+2bc+c^2\\b^2=a^2+2ac+c^2\\c^2=a^2+2ab+b^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-b^2-c^2=2bc\\b^2-a^2-c^2=2ac\\c^2-a^2-b^2=2ab\end{matrix}\right.\)

\(\Leftrightarrow Q=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

Trường hợp 2: a=b=c

\(\Leftrightarrow Q=\dfrac{a^2}{a^2-a^2-a^2}+\dfrac{b^2}{b^2-b^2-b^2}+\dfrac{c^2}{c^2-c^2-c^2}\)

\(\Leftrightarrow Q=\dfrac{a^2}{-2a^2}+\dfrac{b^2}{-2b^2}+\dfrac{c^2}{-2c^2}=\dfrac{-1}{2}+\dfrac{-1}{2}+\dfrac{-1}{2}=\dfrac{-3}{2}\)

\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

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

\(C=ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b-a+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b\right)+bc\left(a-c\right)+ac\left(a-c\right)=b\left(a+b\right)\left(a-c\right)+c\left(a-c\right)\left(a+b\right)=\left(a+b\right)\left(c+c\right)\left(a-c\right)\)

Có:

\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)

Thay \(a=b=c\) vào \(A\), ta được:

\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)

\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)

\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)

\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)

\(=\dfrac{3}{2017^2}\)

Vậy: ...

a) (a-b)(b-c)(a-c).

b) (a-b)(b-c)(a - c)(a + b + c).