Calculation of determinant using its properties

Question

The task is to calculate the following determinant by using the properties of a determinant: $$\begin{vmatrix} n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2 \\ \end{vmatrix}$$ Hint: The result is $-8$.

Answer 1

Expanding all the terms isn't so bad in this example.

ROWS-TRANSFORMATIONS $$A=\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ n^2+2n+1&n^2+4n+4&n^2+6n+9\\ n^2+4n+4&n^2+6n+9&n^2+8n+16 \end{pmatrix}\sim \begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ n^2+2n+1&n^2+4n+4&n^2+6n+9\\ 2n+3&2n+5&2n+7 \end{pmatrix}\sim\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ 2n+1&2n+3&2n+5\\ 2n+3&2n+5&2n+7 \end{pmatrix}\sim\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ 2n+1&2n+3&2n+5\\ 2&2&2 \end{pmatrix}$$ COLUMN-TRANSFORMATIONS $$\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ 2n+1&2n+3&2n+5\\ 2&2&2 \end{pmatrix}\sim\begin{pmatrix} n^2&n^2+2n+1&2n+3\\ 2n+1&2n+3&2\\ 2&2&0 \end{pmatrix}\sim\begin{pmatrix} n^2&2n+1&2n+3\\ 2n+1&2&2\\ 2&0&0 \end{pmatrix}\sim\begin{pmatrix} n^2&2n+1&2\\ 2n+1&2&0\\ 2&0&0 \end{pmatrix}$$ So $\mathrm{det}(A)=-8$.