Question

Prove that (a+b+c)³-a³-b³-c³= 3(a+b)(b+c)(c+a). Solve this fast please it's urgent​

Answer 1

Answer:

(a+b+c)³-a³-b³-c³

=a³+b³+c³+3{(a+b)(b+c)(c+a)}-a³-b³-c³

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

Answer 2

We have to prove that (a + b + c)³ - a³ - b³ - c³ = 3(a + b ) (b + c) (c + a).

Considering LHS,

LHS : (a + b + c)³ - a³ - b³ - c³

= [(a + b + c)³ - a³] - (b³ + c³)

Using the algebraic identity,

a³ - b³ = (a - b)(a² + b² + ab)

(a + b + c)³ - a³ = ((a + b + c) - a)((a + b + c)² + a² + (a + b + c)a)

Using the algebraic identity,

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

= (b + c)(a² + b² + c² + 2(ab + bc + ca) + a² + a² + ab + ac)

= (b + c)(3a² + b² + c² + 3ab + 2bc + 3ac)

Using the algebraic identity,

a³ + b³ = (a + b)(a² + b² - ab)

b³ + c³ = (b + c)(b² + c² - bc)

So, [(a + b + c)³ - a³] - (b³ + c³) = (b + c)(3a² + b² + c² + 3ab + 2bc + 3ac) - [(b + c)(b² + c² - bc)]

= (b + c)[3a² + b² + c² + 3ab + 2bc + 3ac - (b² + c² - bc)]

= (b + c)[3a² + b² + c² + 3ab + 2bc + 3ac - b² - c² + bc]

= (b + c)[3a² + 3ab + 3bc + 3ac]

= (b + c)[3(a² + ab + bc + ac)]

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

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

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

= RHS

Hence proved.