# Using properties of determinants, prove that $\begin{bmatrix} 2ab & a^2 & b^2 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{bmatrix} = -(a^3+b^3)^2$

Toolbox:
• Elementary transformation in a determinant can be made by
• a)by interchanging two rows or columns.
• b)by adding or subtracting two or more columns
• The value of the determinant can be determined by expanding it along any rows or columns
Step 1:
Let $\Delta= \begin{bmatrix} 2ab & a^2 & b^2 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{bmatrix}$
Apply $C_1 \to C_1+C_2+C_3$
$\Delta=\begin{bmatrix} a^2+b^2+2ab & a^2+b^2+2ab & a^2+b^2+2ab \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{bmatrix}$
$(a+b)^2=a^2+b^2+2ab$
Take $(a+b)^2$ as common factor from $R_1$
$\Delta=(a+b)^2\begin{bmatrix} 1 & 1 & 1 \\ a^2 & b^2 & 2ab \\ b^2 & 2ab & a^2 \end{bmatrix}$
Apply $C_2 \to C_2-C_3\; and \; C_3 \to C_3-C_1$
$\Delta=(a+b)^2\begin{bmatrix} 1 & 0 & 0 \\ a^2 & b^2-2ab & 2ab-a^2 \\ b^2 & 2ab-a^2 & a^2-b^2 \end{bmatrix}$
Now apply $C_2 \to C_2 +C_3$
$\Delta=(a+b)^2\begin{bmatrix} 1 & 0 & 0 \\ a^2+b^2 & b^2-a^2 & 2ab-a^2 \\ b^2 & 2ab-b^2 & a^2-b^2 \end{bmatrix}$
Step 2:
Now expand along $R_1$
$\Delta=(a+b)^2 \bigg[1(b^2-a^2)(a^2-b^2)-(2ab-a^2)(2ab-b^2)\bigg]$
on simplifying we get
$\Delta=(a+b)^2 \bigg[1(b^2-a^2)(a^2-b^2)-(2ab-a^2)(2ab-b^2)\bigg]$
$=(a+b)^2 \bigg[b^2a^2-b^4-a^4+a^2b^2-(4a^2b^2-2ab^3-2a^3b+a^2b^2)\bigg]$
$=(a+b)^2 \bigg[2a^2b^2-b^4-a^4-4a^2b^2+2ab^3+2a^3b-a^2b^2\bigg]$
$=(a+b)^2 \bigg[-a^4-b^4-3a^2b^2+2ab^3+2a^3b\bigg]$
$(a^2+b^2-ab)^2=a^4+b^4+a^2b^2+2a^2b^2-2b^3a-2a^3b$
using the identity $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Therefore $(a^2+b^2-ab)^2=a^4+b^4+3a^2b^2-2ab^3-2a^3b$
Therefore $-(a^2+b^2-ab)^2=-a^4-b^4-3a^2b^2+2ab^3+2a^3b$
substituting this we get,
Therefore $\Delta=-\bigg[(a+b)(a^2+b^2-ab)\bigg]^2$
$a^3+b^3=(a+b)(a^2+b^2-ab)$
Therefore $\Delta=-[(a^3+b^3)]^2$
Hence proved