Given that $:(a^{2}+b^{2}),(b^{2}+c^{2})$ and $(a^{2}+c^{2})$ are in GP.
If $x,y,z$ are in GP, then $\dfrac{y}{x} = \dfrac{z}{y}\implies y^{2} = xz$
$\therefore (a^{2}+b^{2}),(b^{2}+c^{2}),(a^{2}+c^{2})$ are in GP
$\implies \dfrac{(b^{2}+c^{2})}{(a^{2}+b^{2})} = \dfrac{(a^{2}+c^{2})}{(b^{2}+c^{2})}$
$\implies (b^{2}+c^{2})(b^{2}+c^{2}) = (a^{2}+c^{2}) (a^{2}+b^{2})$
$\implies (b^{2}+c^{2})^{2} = (a^{2}+c^{2}) (a^{2}+b^{2})$
$\implies b^{4}+c^{4} + 2b^{2}c^{2} = a^{4} + a^{2}b^{2} + c^{2}a^{2} + c^{2}b^{2}$
$\implies b^{4}+c^{4} + b^{2}c^{2} = a^{4} + a^{2}b^{2} + a^{2}c^{2}$
$\implies b^{4} + b^{2}c^{2} – a^{2}b^{2} – a^{2}c^{2} = a^{4} – c^{4}$
$\implies b^{2}(b^{2} + c^{2}) – a^{2}(b^{2} + c^{2}) = a^{4} – c^{4}$
$\implies (b^{2} + c^{2})(b^{2}-a^{2}) = a^{4} – c^{4}$
$\implies b^{2}-a^{2} = \dfrac{a^{4} – c^{4}}{b^{2} + c^{2}}$
So, the correct answer is $(B).$