Đáp án + Giải thích các bước giải:

Ta có: $a +b + c = 0$

Suy ra $\begin {cases} a = - b - c \\ b = -c - a \\ c = -a - b \end {cases}$

$\begin{array}{l} &\left(\dfrac{b - c}{a} + \dfrac{c - a}{b} + \dfrac{a - b}{c}\right)\left(\dfrac{a}{b - c} + \dfrac{b}{c - a} + \dfrac{c}{a - b}\right) (a \ne b \ne c, a \ne 0, b \ne 0, c \ne 0)
\\ &= 1 + \dfrac{a}{b - c}\left(\dfrac{c - a}{b} + \dfrac{a - b}{c}\right) + 1 + \dfrac{b}{c - a}\left(\dfrac{b - c}{a} + \dfrac{a - b}{c}\right)
\\ &+ 1 + \dfrac{c}{a - b}\left(\dfrac{b - c}{a} + \dfrac{c - a}{b}\right)\end{array}$

$\begin{array}{l}\textrm{Đặt} &\textbf{A}_1 = 1 + \dfrac{a}{b - c}\left(\dfrac{c - a}{b} + \dfrac{a - b}{c}\right)
\\ & \textbf{A}_2 = 1 + \dfrac{b}{c - a}\left(\dfrac{b - c}{a} + \dfrac{a - b}{c}\right)
\\ &\textbf{A}_3 = 1 + \dfrac{c}{a - b}\left(\dfrac{b - c}{a} + \dfrac{c - a}{b}\right) \end{array}$

Biểu thức ban đầu có giá trị là $\textbf{A}_1 + \textbf{A}_2+ \textbf{A}_3$

Ta có: 

$\begin{array}{l} &\textbf{A}_1 = 1 + \dfrac{a}{b - c}\left(\dfrac{c - a}{b} + \dfrac{a - b}{c}\right)
\\ &= 1 + \dfrac{a}{b - c} \cdot \dfrac{c^2 - ac + ab - b^2}{bc}
\\ &= 1 + \dfrac{a}{b - c} \cdot \dfrac{a(b - c) - (b + c)(b - c)}{bc}
\\ &= 1 + \dfrac{a}{b - c} \cdot \dfrac{(b - c)(a - b - c)}{bc}
\\ &= 1 + \dfrac{a(a - b - c)}{bc}
\\ &= 1 + \dfrac{2a^2}{bc}
\\ &= 1 + \dfrac{2a^3}{abc}\end{array}$

$\begin{array}{l} &\textbf{A}_2 = 1 + \dfrac{b}{c - a}\left(\dfrac{b - c}{a} + \dfrac{a - b}{c}\right)
\\ &= 1 + \dfrac{b}{c - a} \cdot \dfrac{bc - c^2 + a^2 - ab}{ac}
\\ &= 1 + \dfrac{b}{c - a} \cdot \dfrac{b(c - a) - (c + a)(c - a)}{ac}
\\ &= 1 +  \dfrac{b}{c - a} \cdot \dfrac{(c - a)(b - c - a)}{ac}
\\ &= 1 + \dfrac{b(b - c - a)}{ac}
\\ &= 1 + \dfrac{2b^2}{ac}
\\ &= 1 + \dfrac{2b^3}{abc}\end{array}$

$\begin{array}{l} &\textbf{A}_3 = 1 + \dfrac{c}{a - b}\left(\dfrac{b - c}{a} + \dfrac{c - a}{b}\right)
\\ &= 1 + \dfrac{c}{a - b} \cdot \dfrac{b^2 - bc + ac - a^2}{ab}
\\ &= 1 + \dfrac{c}{a - b} \cdot \dfrac{c(a - b) - (a + b)(a - b)}{ab}
\\ &= 1 + \dfrac{c}{a - b} \cdot \dfrac{(a - b)(c - a - b)}{ab}
\\ &= 1 + \dfrac{c(c - a - b)}{ab}
\\ &= 1 + \dfrac{2c^2}{ab}
\\ &= 1 + \dfrac{2c^3}{abc}\end{array}$

Suy ra $\textbf{A}_1 + \textbf{A}_2 + \textbf{A}_3 = 3 + \dfrac{2(a^3 + b^3 + c^3)}{abc}$

Ta có:

$\begin{array}{l} & a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2) - ab^2 - ac^2 - a^2b - c^2b - a^2c - b^2c
\\ &= (a + b + c)(a^2 + b^2 +c^2) - (ab^2 + a^2b + abc) - (ac^2 + a^2c + abc)
\\ &- (bc^2 + b^2c + abc) + 3abc
\\ &= (a + b + c)(a^2 + b^2 + c^2) - ab(a + b + c) - ac(a + b + c)
\\ & - bc(a + b + c)+ 3abc
\\ &= (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 3abc\end{array}$

Do đó:

$\textbf{A}_1 + \textbf{A}_2 + \textbf{A}_3 = 3 + \dfrac{2(a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 6abc}{abc}$
$\begin{array}{l} &= 3 + \dfrac{0(a^2 + b^2 + c^2 - ab - ac - bc) + 6abc}{abc}
\\ &= 3 + \dfrac{6abc}{abc}
\\ &= 9 \end{array}$

Vậy biểu thức có giá trị bằng $9$