If ${\left(1+x+x^2\right)}^{48}=a_0+a_{1}x+$$a_2{x}^2+\dots +a_{96}{x}^{96},$then value of $a_0-a_2+a_4-a_6+\dots$ $+a_{96}$ is

Putting$x = i,$ we get 

$\begin{array}{l}{\left(1+i+i^2\right)}^{48}=a_0+a_{1}i+a_2{i}^2+\dots +a^{98}{i}^{96}\\ i^{48}=\left(a_0-a_2+a_4-\dots \right)+i\left(a_1-a_3+\dots \right)\end{array}$

Equating the real parts we get 

$a_0-a_2+a_4-a_6+\dots +a_{96}=1$