Solving a 'Harvard' University entrance exam question

Added: 2026-05-31

[00:00:00]Hello Friends find the value of 'c' If c+c+c=c.c.c let's have a solution here If we put c=0 you can see 0+0+0 which is 0 and on right hand side, 0.0.0 which is 0 0=0 which shows that c=0 is our right solution but you can see c+c+c on left hand side is 3c and on right hand side c.c=c^2 c^2.c=c^3 3c=c^3 since this has power 3 so It has three solutions 1 solution is c=0 and there are two more solutions to solve this, 3c=c^3 this is same as c^3-3c=0 c common we have c(c^2-3)=0 since 3=(√3)^2 square will be cancel from square root we get 3 we have c(c^2-(√3)^2)=0 as we know that a^2-b^2=(a+b)(a-b) so It will be equal to c[(c+√3)(c-√3)]=0 c1=0 first solution c+√3=0 c2=-√3 second solution c-√3=0 c3=√3 third solution so I'm right here If we have 3 power It has three solutions first solution c1=0 second solution c2=-√3 and third solution c3=√3 now I'm going to verify c+c+c=c.c.c put c1=0 0+0+0=0.0.0 0=0 which is absolutely correct If we put c2=-√3 we have, (-√3)+(-√3)+(-√3)=(-√3).(-√3).(-√3) If we common √3 we have -1-1-1 which is -3 √3(-3) and here minus minus plus plus minus minus If we you see √3.√3 we get 3 because √3.√3 we have (√3)^2 square will be cancel from square root we get 3 I hope you understood √3(-3)=-3√3 -3√3=-3√3 L.H.S=R.H.S which is absolutely correct now take c3=√3 I'm going to put here √3+√3+√3=√3.√3.√3 √3 common we get 3√3 like before √3.√3=3 3√3=3√3 L.H.S=R.H.S which is also absolutely correct so finally c1=0 c2=-√3 c3=√3 three solutions 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 and must also visit the Playlists of this channel to learn more and more ok bye