Prove $3(x+y)(x+z)(y+z)\neq a$ cubic when $x,y,z$ are different co-primal positive integers.
I believe you can look at the prime factors of $x,y$ and $z$. As for the equation to equal a cubic, the prime factors must all be present in multiples of $3$'s.
What you're asking to prove is not always true. One specific counter-example, among infinitely many, is
$$x = 19 \tag{1}\label{eq1A}$$
$$y = 324 = 2^2 \times 3^4 \tag{2}\label{eq2A}$$
$$z = 4\text{,}589 = 13 \times 353 \tag{3}\label{eq3A}$$
These are all distinct, co-primal positive integers. You next have
$$x + y = 343 = 7^3 \tag{4}\label{eq4A}$$
$$x + z = 4\text{,}608 = 3^2 \times 2^9 \tag{5}\label{eq5A}$$
$$y + z = 4\text{,}913 = 17^3 \tag{6}\label{eq6A}$$
This gives
$$\begin{equation}\begin{aligned} 3(x + y)(x + z)(y + z) & = 3(343)(4\text{,}608)(4\text{,}913) \\ & = 3(7^3 \times (3^2 \times 2^9) \times 17^3) \\ & = (2^3)^3 \times 3^3 \times 7^3 \times 17^3 \\ & = (8 \times 3 \times 7 \times 17)^3 \\ & = (2\text{,}856)^3 \end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Are there perhaps some other required conditions which were not mentioned?