Would this be a legal move in a Matrix? Here I have a matrix:
\begin{pmatrix}
    b^2c^2 & bc & b+c \\
    c^2a^2 & ca & c+a \\
    a^2b^2 & ab & a+b \\
    \end{pmatrix}
I was wondering can I do the following operation on this:
$C2 \to C2 \times C2$
So it would give me the following:
\begin{pmatrix}
    b^2c^2 & b^2c^2 & b+c \\
    c^2a^2 & c^2a^2 & c+a \\
    a^2b^2 & a^2b^2 & a+b \\
    \end{pmatrix}
This is needed in order to prove the determinant is equal to zero.

Since this will (for most $a$ and $b$) change the value of the determinant, lets use co-factor multiplication by your matrix $M$ to prove this, which shouldn't be too bad given it is a $3x3$ matrix. I will denote each $2x2$ co factor as $A_i$ for simplicity.
$$det(M) = (b+c)det(A_1) - (c+a)det(A_2) + (a+b)det(A_3)$$
$$det(M) = (b+c)(a^3c^2b-a^3b^2c)-(c+a)(c^3b^2a-c^3a^2b)+(a+b)(b^3c^2a -b^3a^2c)$$
Hmm... maybe this wasn't the right approach but whatever too late now
$$det(M) = (a^3c^2b^2-a^3b^3c+a^3c^3b-a^3b^2c^2)-(b^3c^3a-b^3a^2c^2+b^3c^2a^2-b^3a^3c)+(c^3b^2a^2-c^3a^3b+c^3b^3a-c^3a^2b^2)$$
Jesus christ... notice that we can take out a $a^3, b^3$ or $c^3$
$$det(M) = a^3(c^2b^2-b^3c+c^3b-b^2c^2)-b^3(c^3a-a^2c^2+c^2a^2-a^3c)+c^3(b^2a^2-a^3b+b^3a-a^2b^2)$$
And then do even better by noticing we can take out a $ab, bc$ or $ac$
$$det(M) = a^3bc(cb-b^2+c^2-bc)-b^3ca(c^2-ac+ca-a^2)+c^3ba(ba-a^2+b^2-ab)$$
Now some terms cancel... thank god....
$$det(M) = a^3bc(-b^2+c^2)-b^3ca(c^2-a^2)+c^3ba(-a^2+b^2)$$
Rearrange some more..
$$det(M) = -a^3bc(b^2-c^2)-b^3ca(c^2-a^2)-c^3ba(a^2-b^2)$$
Factor out a $-abc$
$$det(M) = -abc(a^2(b^2-c^2)+b^2(c^2-a^2)+c^2(a^2-b^2))$$
Distribute
$$det(M) = -abc((a^2b^2-a^2c^2)+(b^2c^2-b^2a^2)+(c^2a^2-c^2b^2))$$
Reorder terms
$$det(M) = -abc(a^2b^2-b^2a^2-a^2c^2+c^2a^2+b^2c^2-c^2b^2)$$
Thus
$$det(M) = -abc(0)$$
$$det(M) = 0$$
I retract my former statement that this wouldn't be too bad, but luckily I denoted the co factors as $A_i$, saving me loads of time. 
