Evaluate ${[a + b + c]}^{3} - a^{3} - b^{3} - c^{3}$

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Solution

Consider ${[a + b + c]}^{3} - a^{3} - b^{3} - c^{3}$
From the algebraic identity, we know that
${(x + y)}^{3} = x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2} . . . . . . . . . . (i)$
Let us assume that
$x = (a + b), y = c$
Now, substituting the values of $x a n d y$ back to $(i)$ we get
$\begin{array}{rcl} {[a + b + c]}^{3} - a^{3} - b^{3} - c^{3} & = & {(a + b)}^{3} + c^{3} + 3 {(a + b)}^{2} c + 3 (a + b) c^{2} - a^{3} - b^{3} - c^{3} . . . . . . . . . . (i i) \\ = & a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} + c^{3} + [3 (a + b) c] [(a + b) + c] - a^{3} - b^{3} - c^{3} \\ = & 3 a^{2} b + 3 a b^{2} + [3 (a + b) c] [(a + b) + c] \\ = & 3 a b (a + b) + [3 (a + b) c] [(a + b) + c] \\ = & 3 (a + b) [a b + c [(a + b) + c]] \\ = & 3 (a + b) [a b + c [a + b + c]] \\ = & 3 (a + b) [a b + a c + b c + c^{2}] \\ = & 3 (a + b) [a (b + c) + c (b + c)] \\ = & 3 (a + b) [(b + c) (a + c)] \\ = & 3 (a + b) (b + c) (a + c) \end{array}$
Hence, ${[a + b + c]}^{3} - a^{3} - b^{3} - c^{3}$ is $3 (a + b) [(b + c) (a + c)]$

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$(a - b)^{3} + (b - c)^{3} + (c - a)^{3} =$

Evaluate $\frac{a^{3} + b^{3} + c^{3} - 3 a b c}{(a + b + c)}, where (a + b + c) \neq 0$ .

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = [{(a + b + c)}^{3} - a^{3}] - [b^{3} + c^{3}]

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

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