Let the reaction be, (P)→(X)+(Y)+(Z)
Let mass of each daughter nuclei be m, then mass defect can be written as ΔM=M−3m.
Now the energy released will be in the form of kinetic energy of the daughter nuclei. Therefore,
ΔMc2=21mV2+21mV2+21mV2⇒V=3m2ΔMc2=M−ΔM2ΔMc2≈cM2ΔM