Which of the following values of a satisfy the equation $\begin{vmatrix}(1+α)^2&(1+2α)^2&(1+3α)^2\\(2+ α)^2&(2+2α)^2&(2+3α)^2\\(3+α)^2&(3+2α)^2&(3+3α)^2\end{vmatrix}=-648α$?

±9

We have,

$\begin{vmatrix}(1+α)^2&(1+2α)^2&(1+3α)^2\\(2+ α)^2&(2+2α)^2&(2+3α)^2\\(3+α)^2&(3+2α)^2&(3+3α)^2\end{vmatrix}=-648α$

$⇒\begin{vmatrix}(1+α)^2&(1+2α)^2&(1+3α)^2\\3+2α &3+4α &3+6α\\2(4+2α) &2(4+4α) &2(4+6α)\end{vmatrix}=-648α$  [Applying $R_2 → R_2-R_1, R_3→ R_3 -R_1$]

$⇒2\begin{vmatrix}(1+α)^2&(1+2α)^2&(1+3α)^2\\3+2α &3+4α &3+6α\\4+2α&4+4α&4+6α\end{vmatrix}=-648α$  [Taking 2 common from $R_3$]

$⇒2\begin{vmatrix}(1+α)^2&(1+2α)^2&(1+3α)^2\\3+2α &3+4α &3+6α\\1&1&1\end{vmatrix}=-648α$  [Applying $R_3 → R_3-R_2$]

$⇒2\begin{vmatrix}(1+α)^2&α(2+3α) &2α (2+4α)\\3+2α &2α &4α\\1&0&0\end{vmatrix}=-648α$  [Applying $C_2 →C_2-C_1, C_3 →C_3-C_1$]

$⇒2\begin{vmatrix}α(2+3α) &2α (2+4α)\\2α &4α\end{vmatrix}=-648α$  [Expanding along $R_3$]

$⇒4α^2\begin{vmatrix}2+3α&2+4α\\2&2\end{vmatrix}=-648α$

$⇒4α^2(4+6α-4-8α)=-648α$

$⇒-8α^3=-648α$

$⇒-8α(α^2-81)=0$

$⇒α=0,α=± 9$