Prove that: | x a a | = (x-a)²(x+2a) | a x a | | a a x |

Steps to Solve

1. Calculate the Determinant

We have the matrix

$$ \begin{vmatrix} x & a & a \ a & x & a \ a & a & x \end{vmatrix} $$

To compute this determinant, we can use the formula for a $3 \times 3$ determinant:

$$ \text{det}(A) = a(ei-fh) - b(di-fg) + c(dh-eg) $$

where the matrix is

$$ \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} $$.

Substituting our values, we get:

• $A = x$, $B = a$, $C = a$, $D = a$, $E = x$, $F = a$, $G = a$, $H = a$, $I = x$.

Thus,

$$ \text{det} = x(x^2 - a^2) - a(a^2 - ax) + a(a^2 - ax) $$.

This simplifies further once we calculate each term.

2. Simplify the Determinant Expression

Let’s calculate the determinant explicitly:

[ \text{det} = x(x^2 - a^2) - a(a^2 - ax) + a(a^2 - ax) ]

Notice that the last two terms cancel each other:

$$ x(x^2 - a^2) = x^3 - xa^2 $$

Next, factor out the $ (x-a) $ term from $ x^2 - a^2 $:

$$ x^3 - xa^2 = (x - a)(x^2 + xa + a^2) $$.

3. Factor Out $(x - a)^2$ from the Simplified Expression

From $ x^2 + xa + a^2 $, we can check if it can be expressed as a product.

Note that:

$$ x^2 + a^2 + ax = (x - a)^2 + 3a^2 $$

Substituting gives

$$ x^2 + xa + a^2 = (x-a)^2 + 3a^2 $$.

Now we include a factor of $ (x - a) $ to obtain the expression for the determinant:

$$ \text{det} = (x - a)^2(x + 2a). $$

4. Final Verification

Check if we obtained the desired form:

$$ \begin{vmatrix} x & a & a \ a & x & a \ a & a & x \end{vmatrix} = (x - a)^2(x + 2a). $$

This verifies that we have proven the equality.