\(a)\) Ta có : 

\(a+b+c=0\)

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

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

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

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

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

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

Vậy \(a^3+b^3+c^3=3abc\)

Chúc bạn học tốt ~ 

a, a+b+c=0 => a+b=-c 

=>(a+b)³=(-c)³

=>a³+3a²b+3ab²+b³=-c³ 

=>a³+3ab(a+b)+b³=-c³

Mà a+b=-c

=>a³-3abc+b³=-c³

=>a³+b³+c³=3abc (đpcm)

b, \(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}\)

mà a³+b³+c³=3abc (bài a)

\(\Rightarrow P=\frac{3abc}{abc}=3\)

Vậy P=3

1)

$2x=a+b+c$

$\Rightarrow x-a=\dfrac{b+c-a}{2},\quad x-b=\dfrac{c+a-b}{2},\quad x-c=\dfrac{a+b-c}{2}$

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$

$=\dfrac{(b+c-a)(c+a-b)+(c+a-b)(a+b-c)+(a+b-c)(b+c-a)}{4}$

$=\dfrac{2ab+2bc-a^2-b^2-c^2+2bc+2ca-a^2-b^2-c^2+2ca+2ab-a^2-b^2-c^2}{4}$

$=\dfrac{4ab+4bc+4ca-3(a^2+b^2+c^2)}{4}$

$=\dfrac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$

$=\dfrac{(2x)^2-2(a^2+b^2+c^2)}{4}$

$=ab+bc+ca-x^2$

$\therefore\ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+ac+bc-x^2.$

2)

$ab+bc+ca=abc,\quad a+b+c=1$

$(a-1)(b-1)(c-1)$$=abc-ab-bc-ca+a+b+c-1$$=abc-ab-bc-ca+1-1$$=abc-(ab+bc+ca)$$=abc-abc$$=0$

$\therefore\ (a-1)(b-1)(c-1)=0.$

a2+b2+c2=1a2+b2+c2=1

|a|;|b|;|c|≤1|a|;|b|;|c|≤1

−1≤a;b;c≤1−1≤a;b;c≤1

(a+1)(b+1)(c+1)≥0(a+1)(b+1)(c+1)≥0

ab+bc+ac+a+b+c+1+abc≥0(1)ab+bc+ac+a+b+c+1+abc≥0(1)

Mặt khác ta có :

(1+a+b+c)2≥0(1+a+b+c)2≥0

a2+b2+c2+2(ab+bc+ac)+2(a+b+c)+1≥0a2+b2+c2+2(ab+bc+ac)+2(a+b+c)+1≥0

2(a+b+c+ab+bc+ac+1)≥02(a+b+c+ab+bc+ac+1)≥0

(a+b+c+ab+bc+ac+1)≥0(2)(a+b+c+ab+bc+ac+1)≥0(2)

trong nâng cao và phát triển có bài này thật đấy

Ta có : \(a^2+b^2+c^2=1\Rightarrow\left|a\right|;\left|b\right|;\left|c\right|\le1\)

\(\Rightarrow-1\le a;b;c\le1\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge0\)

\(\Rightarrow a+b+c+ab+ac+bc+abc+1\ge0\left(1\right)\)

Lại có : \(\left(a+b+c+1\right)^2\ge0\)

\(\Leftrightarrow\left(a+b+c\right)^2+2\left(a+b+c\right)+1\ge0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac+a+b+c\right)+1\ge0\)

\(\Leftrightarrow2\left(ab+bc+ac+a+b+c+1\right)\ge0\)

\(\Leftrightarrow ab+bc+ac+a+b+c+1\ge0\left(2\right)\)

Từ ( 1 ) ; ( 2 ) \(\Rightarrow abc+2\left(ab+bc+ac+a+b+c+1\right)\ge0\left(đpcm\right)\)