Determinant Proof: Show that a Given Determinant Equals $2abc(a+b+c)^3$
Problem: Show that
$$ \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = 2abc(a+b+c)^3 $$
Step-by-Step Solution
Step 1: Define the Determinant
Let the given determinant be:
$$ \Delta = \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} $$
Step 2: Apply Column Operations
Apply: $$ C_1 \rightarrow C_1 – C_3, \quad C_2 \rightarrow C_2 – C_3 $$
$$ \Delta = \begin{vmatrix} (b+c)^2 – a^2 & 0 & a^2 \\ 0 & (c+a)^2 – b^2 & b^2 \\ c^2 – (a+b)^2 & c^2 – (a+b)^2 & (a+b)^2 \end{vmatrix} $$
Step 3: Use Identity
Using the identity: $$ x^2 – y^2 = (x+y)(x-y) $$
$$ \Delta = \begin{vmatrix} (a+b+c)(b+c-a) & 0 & a^2 \\ 0 & (a+b+c)(c+a-b) & b^2 \\ (a+b+c)(c-a-b) & (a+b+c)(c-a-b) & (a+b)^2 \end{vmatrix} $$
Step 4: Factor Out Common Term
Factor $(a+b+c)$ from $C_1$ and $C_2$:
$$ \Delta = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ c-a-b & c-a-b & (a+b)^2 \end{vmatrix} $$
Step 5: Apply Row Operation
Apply: $$ R_3 \rightarrow R_3 – (R_1 + R_2) $$
$$ \Delta = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ -2b & -2a & 2ab \end{vmatrix} $$
Step 6: Apply Column Transformations
Apply: $$ C_1 \rightarrow C_1 + \frac{1}{a}C_3, \quad C_2 \rightarrow C_2 + \frac{1}{b}C_3 $$
$$ \Delta = (a+b+c)^2 \begin{vmatrix} b+c & \frac{a^2}{b} & a^2 \\ \frac{b^2}{a} & c+a & b^2 \\ 0 & 0 & 2ab \end{vmatrix} $$
Step 7: Expand the Determinant
Expanding along the third row:
$$ \Delta = (a+b+c)^2 \cdot 2ab \left[ (b+c)(c+a) – \left(\frac{a^2}{b} \cdot \frac{b^2}{a}\right) \right] $$
$$ = (a+b+c)^2 \cdot 2ab [ bc + ab + c^2 + ac – ab ] $$
$$ = (a+b+c)^2 \cdot 2ab [ c(a+b+c) ] $$
Final Result
$$ \Delta = 2abc(a+b+c)^3 $$