Evaluate [a + b + c]³ - a³ - b³ - c³

To find the value of [a + b + c]³ - a³ - b³ - c³ we will use algebraic identities.

Answer: [a + b + c]³ - a³ - b³ - c³ = 3 (a + b)(b + c)(c + a).

Let's look into the detailed solution.

Explanation: 

Consider the algebraic identity below,

(x + y)³ = x³ + y³ + 3x² y + 3xy² -------------- (1)

Let x = (a + b) and y = c.

Now, substituting the values of x and y back to (1) and taking (- a³ - b³ - c³) on both the sides, we get,

[a + b + c]³ - a³ - b³ - c³ = (a + b)³ + c³ + 3(a + b)² c + 3(a + b) c² - a³ - b³ - c³  ---------------- (2)

Now, we know that (a + b)³ = a³ + b³ + 3a² b + 3ab² 

On further expanding (2) and by taking 3(a + b)c common, we get,

= a³ + b³ + 3a² b + 3ab² + c³ + 3(a + b)c × {(a + b) + c} - a³ - b³ - c³

= 3a² b + 3ab² + 3(a + b)c × {(a + b) + c}

Taking 3ab common, we can simplify it as,

= 3ab(a + b) + 3(a + b)c × {(a + b) + c}

= 3(a + b) × {ab + ac + c² + bc}

= 3(a + b) × {a(b + c) + c(c + b)}   

= 3(a + b)(b + c)(a + c)

So, the value of [a +b +c ]³ - a³ - b³ - c³ is 3(a + b)(b + c)(a + c).