Let $\begin{vmatrix}1 + x&x&x^2\\x&1 + x&x^2\\x^2&x&1+x\end{vmatrix}=ax^5 + bx^4 + cx^3 + dx^2 + λx + μ$ be an identity in x, where $a, b, c, d, λ, μ$ are independent of x. Then, the value of λ is

3

Let $Δ=\begin{vmatrix}1 + x&x&x^2\\x&1 + x&x^2\\x^2&x&1+x\end{vmatrix}$

Applying $C_1→C_1 + C_2 + C_3$, we get

$Δ=(1+x)^2\begin{vmatrix}1&x&x^2\\1&1 + x&x^2\\1&x&1+x\end{vmatrix}$

Clearly, $λ=(\frac{dΔ}{dx})_{x=0}$

We have,

$\frac{dΔ}{dx}=2(1+x)\begin{vmatrix}1&x&x^2\\1&1 + x&x^2\\1&x&1+x\end{vmatrix}+(1+x)^2\begin{vmatrix}1&1&x^2\\1&1&x^2\\1&1&1+x\end{vmatrix}+(1+x)^2\begin{vmatrix}1&x&2x\\1&1 + x&2x\\1&x&1\end{vmatrix}$

$∴(\frac{dΔ}{dx})_{x=0}=2\begin{vmatrix}1&0&0\\1&1&0\\1&0&1\end{vmatrix}+0+\begin{vmatrix}1&0&0\\1&1&0\\1&0&1\end{vmatrix}=2+1=3$

Hence, $λ=3$.