Equivalent Cube Like Equivalent Resistance

I will use this identity─ $(a+b+c)^3 = a^3+b^3+c^3+3(a+b)(b+c)(c+a)$

As stated in the problem statement, $(a+b+c)^3 = a^3+b^3+c^3$ , so we can say─

$0 = 3(a+b)(b+c)(c+a)$

$\implies 0 = (a+b)(b+c)(c+a)$

$\implies a+b = 0 \text{ or } b+c=0 \text{ or } c+a=0$

$\implies a= -b \text{ or } b=-c \text{ or } c=-a$ , which is a contradiction whenever $a, b$ and $c$ are positive integers.

So, $\boxed{\text{False}}$ is the answer.

$(a+b+c)^3=a^3+b^3+c^3\\ a^3+b^3+c^3+3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc=a^3+b^3+c^3\\ a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+2abc=0\\ a^2c+2abc+b^2c+a^2b+ab^2+ac^2+bc^2=0\\ c(a^2+2ab+b^2)+ab(a+b)+c^2(a+b)=0\\ c(a+b)^2+(ab+c^2)(a+b)=0\\ (c(a+b)+ab+c^2)(a+b)=0\\ (ab+ac+bc+c^2)(a+b)=0\\ (a(b+c)+c(b+c))(a+b)=0\\ (a+c)(b+c)(a+b)=0$ Therefore either $a=-b$ , $a=-c$ or $b=-c$ . However, a, b and c are all positive integers, so no triplet exists.