Benedictus qui venit in nomine Domini.
Oh ... . I’m sorry very much ... .

Really, this problem is not a homework problem ... . I just want know the brief clue of alternative solution to solve the equation ... . I just know that
$|\mathbf{V}|^2={V_x}^2+{V_y}^2+{V_z}^2$ ,
$\Gamma=[1-({V_x}^2+{V_y}^2+{V_z}^2)/c^2]^{-1/2}$ ,
$\mathbf{v}_A\cdot\mathbf{V}=v_{Ax}V_x+v_{Ay}V_y+v_{Az}V_z$ , and
$\mathbf{v}_B\cdot\mathbf{V}=v_{Bx}V_x+v_{By}V_y+v_{Bz}V_z$ ,
in Cartesian coordinate system ... .
Maybe ... , its analytic solution does not exist ... . If in special case (one-dimension), $\mathbf{v}_A=v_A\mathbf{i}$ , $\mathbf{v}_B=v_B\mathbf{i}$ , and $\mathbf{V}=V\mathbf{i}$ ... , then the vector equation becomes
$\displaystyle\frac{V^2(v_A-\Gamma{V})+(\Gamma-1)v_AV^2}{c^2-v_AV}=\frac{V^2(\Gamma{V}-v_B)+(1-\Gamma)v_BV^2}{c^2-v_BV}$ ,
alias $\displaystyle\frac{v_A-V}{c^2-v_AV}=\frac{V-v_B}{c^2-v_BV}$ ,
alias $c^2v_A-(v_Av_B+c^2)V+v_BV^2=-c^2v_B+(v_Av_B+c^2)V-v_AV^2$ ,
alias $(v_A+v_B)V^2-2(v_Av_B+c^2)V+c^2(v_A+v_B)=0$ ,
alias $\displaystyle\,V=\frac{v_Av_B+c^2\pm\sqrt{{v_A}^2{v_B}^2+c^2(c^2-{v_A}^2-{v_{B}}^2)}}{v_A+v_B}$ ... .
alias $\displaystyle\,V=\frac{v_Av_B+c^2\pm\sqrt{c^2-{v_A}^2}\sqrt{c^2-{v_B}^2}}{v_A+v_B}$ ... .
But ... , this is a special case in one-dimension, not 3-dimension ... .
Dalam Nama Bapa dan Putera dan Roh Kudus. Amin.