This is a question from 4U maths (highest level of Y12 maths in Australia) from a generally difficult paper. The question itself does not define what a, b, and c are - based on the comments, we assume a, b, c are real and non-negative.
It is suggested, but not required, to use the following (already proven in earlier part) identity: $a^3+b^3+c^3\ge3abc$
I have tried various ways to solve this, including moving everything to one side and proving that it must be positive, and using AM $\ge$ GM scenarios, as well as multiple applications/substitutions of the proved identity, but did not come up with a valid solution.
I have seen the solution, which uses the identity proved 4 times, substituting a,b,c with variations of $(\frac{a^3}{1+a^3})^\frac{1}{3}, (\frac{b^3}{1+b^3})^\frac{1}{3}, (\frac{c^3}{1+c^3})^\frac{1}{3}, \frac{1}{(1+a^3)^\frac{1}{3}}$ etc. and working from there (summing and algebra) to reach the required result.
I would like to know if that is any other way we can reach the required result other than the method used by the solution.
Original: This is not an answer. I do not have enough reputation to comment so I leave it here.
Is there any typo in the problem? If no, this inequality is not correct in general. For example take $a=-1,b=c=3$. Then $LHS=0$ and $RHS=8$. Note that this counterexample shows $(1+a^3)(1+b^3)(1+c^3)\geq\frac{(ab+bc+ca+1)^3}{8}\cdots(*)$ is not true. I guess maybe the right question is to prove (*) for $a,b,c\geq0$?
Edit:
If the question is $(1+a)^3(1+b)^3(1+c)^3\geq\frac{(ab+bc+ca+1)^3}{8}$, then Prism's comment already solves this problem. It is clear that the equality never holds under the condition $a,b,c\geq 0$.
Now we prove (*). We will use $\sum$ to denote the cyclic sum, i.e. $$\sum f(a,b,c):=f(a,b,c)+f(b,c,a)+f(c,a,b).$$ By direct calculation it is equivalent to show $$8(1+\sum a^3+\sum a^3b^3+a^3b^3c^3)\geq(\sum ab)^3+3(\sum ab)^2+3(\sum ab)+1.$$ One can check that (by direct calculation again) $$\begin{aligned}RHS&=\sum a^3b^3+3\sum ab(a^2c^2+b^2c^2)+6a^2b^2c^2+3\sum a^2b^2+3\sum ab(ac+bc)+3\sum ab+1\\ &=\sum a^3b^3+3abc\sum ab(a+b)+6a^2b^2c^2+3\sum a^2b^2+6abc\sum a+3\sum ab+1\end{aligned}.$$ Now we have
(1) $2\sum a^3b^3+3=\sum (a^3b^3+a^3b^3+1)\geq\sum 3a^2b^2$ by AM-GM.
(2) $6(abc)^3+2\sum a^3b^3+2\sum a^3\geq3abc\sum ab(a+b)$ because:
$\;\;\;$(2a) $\sum((abc)^3+a^3b^3+a^3)\geq\sum(3abc\cdot a^2b)$ by AM-GM.
$\;\;\;$(2b) $\sum((abc)^3+a^3b^3+b^3)\geq\sum(3abc\cdot ab^2)$ by AM-GM.
(3) $2\sum a^3b^3+4\sum a^3\geq 6abc\sum a$ because.
$\;\;\;$(3a) $\sum(a^3b^3+a^3+c^3)\geq\sum(3abc\cdot a)$ by AM-GM.
$\;\;\;$(3b) $\sum(a^3b^3+b^3+c^3)\geq\sum(3abc\cdot b)$ by AM-GM.
(4) $\sum(a^3b^3)\geq 3a^2b^2c^2$ by AM-GM.
(5) $2(abc)^3+1\geq3a^2b^2c^2$ by AM-GM.
(6) $2\sum a^3+3=\sum(a^3+b^3+1)\geq 3\sum ab$ by AM-GM.
Taking the sum of (1) to (6) gives us the desired inequality. One can see that the equality holds if and only if $a=b=c=1$.