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

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

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