Question

$\begin{vmatrix} a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2 \end{vmatrix} = (a^3 + b^3)^2$

Answer 1

$(a^3 + b^3)^2$

Answer 2

Let the given determinant be $D$.

$D = \begin{vmatrix} a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2 \end{vmatrix}$

This is a cyclic determinant of the form $\begin{vmatrix} x & y & z \\ z & x & y \\ y & z & x \end{vmatrix}$, where $x=a^2$, $y=2ab$, and $z=b^2$.

The value of a $3 \times 3$ cyclic determinant is given by the formula:

$\begin{vmatrix} x & y & z \\ z & x & y \\ y & z & x \end{vmatrix} = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.

Substitute the values of $x, y, z$:

$x+y+z = a^2 + 2ab + b^2 = (a+b)^2$.

$x^2 = (a^2)^2 = a^4$ $y^2 = (2ab)^2 = 4a^2b^2$ $z^2 = (b^2)^2 = b^4$ $xy = a^2(2ab) = 2a^3b$ $yz = (2ab)(b^2) = 2ab^3$ $zx = b^2(a^2) = a^2b^2$

Now calculate $x^2+y^2+z^2-xy-yz-zx$:

$a^4 + 4a^2b^2 + b^4 - 2a^3b - 2ab^3 - a^2b^2$

$= a^4 - 2a^3b + (4a^2b^2 - a^2b^2) - 2ab^3 + b^4$

$= a^4 - 2a^3b + 3a^2b^2 - 2ab^3 + b^4$.

This expression is the expansion of $(a^2 - ab + b^2)^2$:

$(a^2 - ab + b^2)^2 = ((a^2+b^2) - ab)^2 = (a^2+b^2)^2 - 2(a^2+b^2)(ab) + (ab)^2$

$= (a^4 + 2a^2b^2 + b^4) - (2a^3b + 2ab^3) + a^2b^2$

$= a^4 + 2a^2b^2 + b^4 - 2a^3b - 2ab^3 + a^2b^2$

$= a^4 - 2a^3b + 3a^2b^2 - 2ab^3 + b^4$.

So, $x^2+y^2+z^2-xy-yz-zx = (a^2 - ab + b^2)^2$.

Now, substitute these back into the determinant formula:

$D = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

$D = (a+b)^2 (a^2 - ab + b^2)^2$

$D = [(a+b)(a^2 - ab + b^2)]^2$

Recall the sum of cubes formula: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

So, $D = (a^3 + b^3)^2$.

The given statement is $\begin{vmatrix} a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2 \end{vmatrix} = (a^3 + b^3)^2$.

Our calculation shows that the determinant is indeed equal to $(a^3 + b^3)^2$.

Thus, the statement is true.