Is the following $3\times 3$ matrix with rational entries invertible? Assume that $(a,b,c)\neq(0,0,0)$ where $a$,$b$, and $c$ are rational numbers. Is the $3\times 3$ matrix
$$\left(\begin{matrix}
                    a & 2c & 2b \\
                    b & a & 2c \\
                    c & b & a
                  \end{matrix}\right)$$
always invertible? The determinant worked out to be $a³-6abc+2b³+4c³$ which doesn't seem to be very illuminating. I tried considering three cases $a\neq 0$, $b\neq 0$, and $c\neq 0$, but am not getting anywhere! Thanks.

Let $d$ be the common denominator of these rational numbers.  If your original matrix were non-invertible then so too would be $d$ times the matrix and all entries would be integers.  Similarly, we can divide by their greatest common divisor while keeping the matrix non-invertible.
We may as well assume then that $a,b,c$ are all integers with at least one non-zero such that $\gcd(a,b,c)=1$.  So, the question becomes, other than the trivial solution, if there exist integers $a,b,c$ with $\gcd(a,b,c)=1$ such that $a^3+2b^3+4c^3-6abc=0$.
If such a solution exists, then $a$ must be even because otherwise with $a$ odd we have an odd number plus or minus multiple even numbers equaling zero, an impossibility.
So... replace $a$ by $2x$ with $x$ an integer.  We have $8x^3 + 2b^3+4c^3-12xbc=0$.  Divide everything by two.  $4x^3+b^3+2c^3-6xbc=0$ which is suspiciously familiar to the original equation.
By the same logic, $b$ is also even.  Replacing $b$ by $2y$ and doing the same thing also shows that $c$ is even.
This is all a contradiction however as it implies that $2$ is a common divisor of $a,b,c$ contradicting that $\gcd(a,b,c)=1$
