$$\begin{vmatrix}
a^2+\lambda^2 & ab+c\lambda & ca-b\lambda \\
ab-c\lambda & b^2+\lambda^2& bc+a\lambda\\
ca+b\lambda & bc-a\lambda & c^2+\lambda^2
\end{vmatrix}.\begin{vmatrix}
\lambda & c & -b\\
-c& \lambda & a\\
b & -a & \lambda
\end{vmatrix}=(1+a^2+b^2+c^2)^3.$$
Then value of $\lambda$ is
options:: (a)$\; 8\;\;$ (b) $\; 27\;\;$ (c)$ \; 1\;\;$ (d) $\; -1\;\;$
actually as I have seen the question. Then I used Multiply these two determinant. but This is very tidious task.
So I want a better Method by which we can easily calculate value of $\lambda$.
So please explain me in detail.
Best Answer
I am not sure this is the best way, but brute force leads to:
$$\lambda^3(\lambda^2 + a^2 + b^2 + c^2)^3 = (1+a^2+b^2+c^2)^3$$
It is very clear what $\lambda$ has to be.
The first determinant yields: $\lambda^2(a^2 + b^2 + c^2 + \lambda^2)^2.$
The second determinant yields: $\lambda(a^2 + b^2 + c^2 + \lambda^2).$
Maybe there is an easy way to take advantage of the second determinant with the RHS, but I am tired and do not see it.