Q.If $x + y + z = 0$, prove that $\begin{vmatrix} xa & yb & zc \\ yc & za & xb \\ zb & xc & ya \end{vmatrix} = xyz \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$.

Concept — expand, then use $x+y+z=0$. Expanding the left determinant along the first row,

Since $x+y+z=0$, the identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ gives $x^3+y^3+z^3=3xyz$. Hence

$\Delta=xyz(a^3+b^3+c^3)-abc(3xyz)=xyz\,(a^3+b^3+c^3-3abc).$ …

The idea

Expand the messy determinant into symmetric pieces. Everything collapses once $x+y+z=0$ is used through the identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.

Step 1 — Expand the left side

Let $\Delta=\begin{vmatrix}xa&yb&zc\\yc&za&xb\\zb&xc&ya\end{vmatrix}$. Expanding along the first row and grouping terms,

Step 2 — Apply $x+y+z=0$

From $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ and $x+y+z=0$,

Therefore

$\Delta=xyz(a^3+b^3+c^3)-abc(3xyz)=xyz\,(a^3+b^3+c^3-3abc).$ …

Method: Expanding a Determinant Identity and Applying a Symmetric-Function Constraint

Use this method whenever you must prove one determinant equals a scalar multiple of another, given an algebraic constraint (like $x+y+z=0$) on some of the variables involved.

Steps

Step 1: Expand the more complicated determinant fully

Expand the left-hand determinant by cofactor expansion and collect the result into natural groups — typically one group multiplying an expression like $a^3+b^3+c^3$ and another multiplying an expression like $x^3+y^3+z^3$, since the entries mix two sets of variables.

Step 2: Bring in the given constraint via a symmetric-function identity

Recall the standard identity

When the problem states $x+y+z=0$, the right side vanishes, so this identity immediately gives $x^3+y^3+z^3=3xyz$ — substitute this simplification into the expansion from Step 1.

Step 3: Simplify to a single product

After substituting, the expression should collapse into a single product of a simple factor (like $xyz$) and a symmetric expression in the remaining variables (like $a^3+b^3+c^3-3abc$).

Step 4: Evaluate the target determinant on the right-hand side independently …

Common Mistakes

Mistake 1: Expanding the left determinant directly instead of grouping symmetric terms

Why it's wrong: brute-force cofactor expansion of a determinant whose nine entries are all products (like $xa$, $yc$, $zb$) generates many cross terms; without deliberately grouping into $xyz(a^3+b^3+c^3)$ and $abc(x^3+y^3+z^3)$, it's very easy to mis-collect terms and lose the clean structure the identity depends on.

Mistake 2: Using $x^3+y^3+z^3=3xyz$ as if it were always true

Why it's wrong: this simplification only holds because $x+y+z=0$ is given (from the identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$); quoting it as a general algebraic fact without tying it to the given condition is a reasoning gap, and forgetting the condition entirely would make the whole proof invalid. …