IRL_ABooklet

$$\begin{array}{rlrlr}{a}^3+b^3+c^3-3abc& =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)& =a^2+b^2+c^2-ab-bc-ca& =(a+b+c{)}^2-3(ab+bc+ca)& =1-3(ab+bc+ca),\end{array}$$ and $$\begin{array}{rlr}(1-a)(1-b)(1-c)& =1-(a+b+c)+(ab+bc+ca)-abc& =ab+bc+ca-abc.\end{array}$$ Hence $$\begin{array}{rlrl}{a}^3+b^3+c^3& =1+3abc-3(ab+bc+ca)& =1-3(ab+bc+ca-abc)& =1-3(1-a)(1-b)(1-c),\end{array}$$ as claimed.

---

Alternative solution.

$$1=(a+b+c{)}^3=a^3+b^3+c^3+3(a^{2}b+a{b}^2+b^{2}c+b{c}^2+c^{2}a+c{a}^2)+6abc.(1)$$ Next we use $a+b+c=1$ to get $1-a=b+c$, $1-b=c+a$, $1-c=a+b$, which gives $$\begin{array}{rl}{a}^3+b^3+c^3+3(1-a)(1-b)(1-c)& =a^3+b^3+c^3+3(a+b)(b+c)(c+a)\\ & =a^3+b^3+c^3+3(a^{2}b+a{b}^2+b^{2}c+b{c}^2+c^{2}a+c{a}^2)+6abc.(2)\end{array}$$ Finally, we note that the expressions on the right-hand sides of (1) and (2) are the same. This completes the proof.

---

Alternative solution.

$$\begin{array}{rl}& a^3+b^3+c^3+3(1-a)(1-b)(1-c)\\ & =a^3+b^3+(1-a-b{)}^3+3(1-a)(1-b)(a+b)\\ & =a^3+b^3+(1-a{)}^3-3b(1-a{)}^2+3b^2(1-a)-b^3+3(1-a)(1-b)(a+b)\\ & =a^3+(1-a)\left((1-a{)}^2-3b(1-a)+3b^2+3(1-b)(a+b)\right)\\ & =a^3+(1-a)\left(1-2a+a^2-3b+3ab+3b^2+3a+3b-3ab-3b^2\right)\\ & =a^3+(1-a)\left(1+a+a^2\right)=a^3+1-a^3=1.\end{array}$$