Solving a 'Stanford' University entrance exam | t=?

Ajouté : 2026-05-30

[00:00:00]Hello Friends find the value of 't' If t^6=2^6 let's have a solution It can be written as t^6-2^6=0 which is same as (t^3)^2-(2^3)^2=0 since 3x2=6 as we know that a^2-b^2=(a+b)(a-b) then we have (t^3+2^3)(t^3-2^3)=0 we have two cases here case I is t^3+2^3=0 and case II is t^3-2^3=0 first of all, I'm going to solve case I since a^3+b^3=(a+b)(a^2-ab+b^2) It will be (t+2)(t^2-2t+2^2)=0 (t+2)(t^2-2t+4)=0 either t+2=0 or t^2-2t+4=0 t=-2 which is our first real solution and here, this is like quadratic equation ax^2+bx+c=0 by comparing, a=1, b=-2, c=4 t=-b±√b^2-4ac/2a t=-(-2)±√(-2)^2-4(1)(4)/2(1) t=2±√4-16/2 in the next step t=2±√-12/2 as √-12=√4x3x-1 It can be written as √4.√3.√-1 2√3i where 'i' is from complex numbers so here √-12=2√3i then It will be t=2±2√3i/2 I hope you understood t=2(1±√3i)/2 2 and 2 cancels t=1±√3i which are complex solutions now Case II which is t^3-2^3=0 as we know a^3-b^3=(a-b)(a^2+ab+b^2) It will be (t-2)(t^2+2t+2^2)=0 (t-2)(t^2+2t+4)=0 either t-2=0 or t^2+2t+4=0 t=2 which is also real solutions and here we have a quadratic equation apply quadratic formula, t=-2±√(2)^2-4(1)(4)/2(1) t=-2±√4-16/2 t=-2±√-12/2 as √-12=2√3i t=-2±2√3i/2 t=2(-1±√3i)/2 by common '2' 2 and 2 cancels t=-1±√3i which are also our complex solutions so finally, we have '6' solutions here t1=-2 t2=1+√3i t3=1-√3i t4=2 t5=-1+√3i t6=-1-√3i 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