Question Number 221968 by MathematicalUser2357 last updated on 14/Jun/25
$$\left({a}+{b}+{c}\right)^{\mathrm{3}} \\ $$
Answered by MrGaster last updated on 14/Jun/25
Commented by MathematicalUser2357 last updated on 14/Jun/25
Thanks. I have been waiting for this formula for a long time.
Answered by fantastic last updated on 14/Jun/25
![let a+b be α So (a+b+c)^3 =(α+c)^3 =α^3 +3α^2 c+3αc^2 +c^3 =(a+b)^3 +3(a+b)^2 c+3(a+b)c^2 +c^3 [∵α=a+b] =a^3 +3a^2 b+3ab^2 +b^3 +3c(a^2 +2ab+b^2 )+3ac^2 +3bc^2 +c^3 =a^3 +b^3 +c^3 +3a^2 b+3ab^2 +3a^2 c+3b^2 c+3ac^2 +3bc^2 +6abc](https://www.tinkutara.com/question/Q221980.png)
$${let}\:{a}+{b}\:{be}\:\alpha \\ $$$${So}\:\left({a}+{b}+{c}\right)^{\mathrm{3}} \\ $$$$=\left(\alpha+{c}\right)^{\mathrm{3}} \\ $$$$=\alpha^{\mathrm{3}} +\mathrm{3}\alpha^{\mathrm{2}} {c}+\mathrm{3}\alpha{c}^{\mathrm{2}} +{c}^{\mathrm{3}} \\ $$$$=\left({a}+{b}\right)^{\mathrm{3}} +\mathrm{3}\left({a}+{b}\right)^{\mathrm{2}} {c}+\mathrm{3}\left({a}+{b}\right){c}^{\mathrm{2}} +{c}^{\mathrm{3}} \left[\because\alpha={a}+{b}\right] \\ $$$$={a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} +\mathrm{3}{c}\left({a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \right)+\mathrm{3}{ac}^{\mathrm{2}} +\mathrm{3}{bc}^{\mathrm{2}} +{c}^{\mathrm{3}} \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} {c}+\mathrm{3}{b}^{\mathrm{2}} {c}+\mathrm{3}{ac}^{\mathrm{2}} +\mathrm{3}{bc}^{\mathrm{2}} +\mathrm{6}{abc} \\ $$
Answered by mr W last updated on 14/Jun/25
$$\left({a}+{b}+{c}\right)^{\mathrm{3}} \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}+\mathrm{3}\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right) \\ $$