Question

Prove that
$(a + b + c) {}^{3} - a {}^{3} - b {}^{3 } - c {}^{3} = 3(a + b)(b + c)(c + a)$
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Answer 1

Question:

Prove that (a + b + c)³ - a³ - b³ - c³ = 3(a + b)(b + c)(c + a)

Solution:

(a + b + c)³ - a³ - b³ - c³ = 3(a + b)(b + c)(c + a)

Take LHS

(a + b + c)³ - a³ - b³ - c³

It can be written as

= [ (a + b) + c ]³ - a³ - b³ - c³

Using algebraic identity (x + y)³ = x³ + y³ + 3xy(x + y)

= (a + b)³ + c³ + 3c(a + b)(a + b + c) - a³ - b³ - c³

Again, Using algebraic identity (x + y)³ = x³ + y³ + 3xy(x + y)

= a³ + b³ + 3ab(a + b) + c³ + 3c(a + b)(a + b + c) - a³ - b³ - c³

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

Taking 3(a + b) common in both terms

= 3(a + b) [ ab + c(a + b + c) ]

= 3(a + b)( ab + ac + bc + c² )

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

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

= RHS

Hence proved