Question

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

Answer 1

Hey there!

_____________________________

This is a Well known identity...

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

But not let's just mug up formula, Let's Derive it :)

$(a+b+c)^{3}=(a+b+c)^{2}(a+b+c)$

$(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+a^{2} (b+c)+b^{2} (a+c)+c^{2}(a+b)+2(ab+bc+ac)(a+b+c)$

$a^{3}+b^{3}+c^{3}+3a^{2}(b+c)+3b^{2}(a+c)+3c^{2}(a+b)+6abc$

Hence, 

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

____________________

Thanks!

Hope it Helps!

Have a great day dear! :)

If any further questions, please feel free to use Comments section below!

Answer 2

Hi.

$(a+b+c)^3\\\\= (a+b+c)^2(a+b+c)\\\\= (a+b+c)(a+b+c)(a+b+c)\\\\= [(a+b+c)(a+b+c)](a+b+c)\\\\= [a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2][a+b+c]\\\\= (a^2+b^2+c^2 + 2ab + 2bc + 2ac)(a+b+c)\\\\= (a+b+c)(a^2+b^2+c^2 + 2ab + 2bc + 2ac)\\\\= a^3 + ab^2 + ac^2 + 2a^2b + 2 abc + 2a^2c + a^2b + b^3 + bc^2 + 2ab^2 + 2b^2c + 2abc + a^2c + b^2c + c^3 + 2abc + 2bc^2 + 2ac^2$

$= a^3 + b^3 + c^3 + ab^2 + 2ab^2 + ac^2 + 2ac^2 + 2a^2b + a^2b + 2abc + 2abc + 2abc + 2a^2c + a^2c + bc^2 + 2bc^2 + b^2c + 2b^2c\\\\= \boxed{a^3 + b^3 + c^3 + 3ab^2 + 3ac^2 + 3a^2b + 6abc + 3a^2c + 3bc^2 + 3b^2c}$

Hope it helps!!