Question

$prove \: that: \\ ( a + b) {}^{3} + (b + c) {}^{3} + (c + a) {}^{3} - 3(a + b)(b + c)(c + a) = 2(a {}^{3} + b {}^{3} + c {}^{3} - 3abc)$
$\textbf{NO SPAMMING!}$

Answer 1

Given : (a + b)^3 + (b +c)^3 + (c + a)^3 - 3(a + b)(b + c)(c + a).

= > a^3 + b^3 + 3ab(a + b) + b^3 + c^3 + 3bc(b + c) + c^3 + a^3 + 3ac(a + c) - 3(a + b)(b + c)(c + a)

= > a^3 + b^3 + 3a^2b + 3ab^2 + b^3 + c^3 + 3b^2c + 3bc^2 + c^3 + a^3 + 3a^2c + 3ac^2 - 3(2abc + a^2b + ac^2 + a^2c + ab^2 + b^2c + bc^2)

= > 2a^3 + 2b^3 + 2c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 - 3(2abc + a^2b + ac^2 + a^2c + ab^2 + b^2c + bc^2)

= > 2(a^3 + b^3 + c^3) + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 - 6abc - 3a^2b - 3ac^2 - 3a^2c - 3ab^2 - 3b^2c - 3bc^2

= > 2(a^3 + b^3 + c^3) - 6abc

= > 2(a^3 + b^3 + c^3 - 3abc).

Hope this helps!

Answer 2

hope it may help you dear