Solving a 'Harvard' University entrance exam question

Added: 2026-05-23

[00:00:00]Hello Friends we have a nice math olympiad algebra problem find the value of 't' If t.t.t=t+t+t let's have a solution this question is same as t^1.t^1.t^1=t+t+t on left hand side, base same powers add then It will be t^(1+1+1) and on right hand side t can be common t(1+1+1) t^(1+1+1)=t(1+1+1) t^3=t(3) we have t^3=3t which can be written as t^3-3t=0 which is same as t^2.t-3t=0 t can be common we have t(t^2-3)=0 two cases here case I is t=0 which is the value of 't' and case II is t^2-3=0 simplify this as a=√a^2=(√a)^2 then It will be t^2-(√3)^2=0 as we know a^2-b^2=(a+b)(a-b) we have (t+√3)(t-√3)=0 either t+√3=0 or t-√3=0 we get t=-√3 and t=+√3 which are values of 't' so we have three solutions here t1=0, t2=-√3, t3=+√3 in the next step I'm going to verify t.t.t=t+t+t first of all, take t1=0 0.0.0=0+0+0 0=0 L.H.S=R.H.S now, I take t2=-√3 put here (-√3).(-√3).(-√3)=(-√3)+(-√3)+(-√3) simplify this (-√3).(-√3)=(-√3)^2 (-√3)^2.(-√3)=-√3(1+1+1) by common -√3 square cancels from square root (-1)^2=(-1)(-1)=+1 so It will be 3.(-√3)=-3√3 -3√3=-3√3 L.H.S=R.H.S now, I'm going to t3=+√3 √3.√3.√3=√3+√3+√3 (√3)^2.√3=3√3 3√3=3√3 L.H.S=R.H.S which shows that t1=0, t2=-√3, t3=+√3 which satisfies the equation t.t.t=t+t+t thanks for watching this video please subscribe this channel to get the notification of my new videos and don't forget to share these videos with your classmates and friends so that they also have a benefit of it you can also visit the Playlists of this channel to learn more and more ok bye