`From a^3 + b^3 + c^3 = 3abc`
`=> (a + b)^3 - 3ab(a + b) + c^3 - 3abc = 0`
`<=> [(a + b)^3 + c^3] - [3ab(a + b) + 3abc] = 0`
`<=> (a + b + c)^3 - 3c(a + b)(a + b + c) - 3ab(a + b + c) = 0`
`<=> (a + b + c)[(a + b + c)^2 - 3c(a + b) - 3ab] = 0`
`<=> (a + b + c)(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 3ac - 3bc - 3ab) = 0`
`<=> (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 0`
`but a + b + c ne 0`
`<=> a^2 + b^2 + c^2 - ab - ac - bc = 0`
`<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 0`
`<=> (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (a^2 - 2ac + c^2) = 0`
`<=> (a - b)^2 + (b - c)^2 + (a - c)^2 = 0`
`but (a - b)^2, (b - c)^2, (a - c)^2 >= 0`
`<=>`$\begin{cases}(a - b)^2 = 0\\(b - c)^2 = 0\\(a - c)^2 = 0\end{cases}$
`<=>` $\begin{cases}a - b = 0\\ b - c = 0\\a - c = 0\end{cases}$
`<=>`$\begin{cases}a = b\\b = c\\a = c \end{cases}$
`<=> a = b = c`
`<=>`$\begin{cases}\dfrac{a}{b} = 1\\\dfrac{b}{c} = 1\\\dfrac{c}{a} = 1\end{cases}$
`Expression A become t o:`
`A = (1 + 1)(1 + 1)(1 + 1)`
`= 2. 2. 2`
`= 8`
`Hence A = 8`
