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

追加: 2026-05-10

[00:00:00]Hello Friends find the value of 't' If t^6=9^6 let's have a solution you can see, this problem is a 6th degree equation has highest power of '6' must remember that If we have 6th degree equation it'll always give 6 different solutions let's start this problem can be written as t^6-9^6=0 t^(3x2)-9^(3x2)=0 since 3x2=6 as we know a^mn=(a^m)^n then It will be (t^3)^2-(9^3)^2=0 apply formula x^2-y^2=(x+y)(x-y) we have (t^3+9^3)(t^3-9^3)=0 we have two cases here case I is t^3+9^3=0 and case II is t^3-9^3=0 first of all, solve case I t^3+9^3=0 by this, we know the formula x^3+y^3=(x+y)(x^2-xy+y^2) then we have (t+9)(t^2-9t+9^2)=0 either t+9=0 or t^2-9t+81=0 here, we get the value of 't1=-9' this is our first real solution and here, we have a quadratic equation ax^2+bx+c=0 by comparing, we have a=1, b=-9, c=81 take discriminant d=b^2-4ac (-9)^2-4(1)(81) simplify this 81-324 -243 <0 which means that It will give us complex solution so apply quadratic formula t=(-b±√d)/2a putting values of a, b and c t=-(-9)±√-243/2 here, you can see 243, simplify 243 divisible 3 81 81 divisible by 3 is 27 27 divisible by 3 is 9 9 divisible by 3 is 3 3 divisible by 3 is 1 t=9±√3^2x3^2x3x-1/2 square cancels from square root 3x3=9 t=9±9√3i/2 we have second and third solution now, take case II t^3-9^3=0 x^3-y^3=(x-y)(x^2+xy+y^2) apply this formula (t-9)(t^2+9t+9)=0 either t-9=0 or t^2+9t+81-=0 fourth solution, t4=-9 a=1, b=9, c=81 like before d=b^2-4ac (9)^2-4(1)(81) 81-324 d=-243 <0 It will give complex solution t=-b±√d/2a t=-9±√-243/2 as like before √-243=9√3i so we can write here t=-9±9√3i/2 fifth and sixth solution t=-9±9√3i/2 so we have 6 solutions here it is clear that 6th degree equation must give 6 different solutions thanks for watching this video please subscribe this channel to get the notification of my new videos ok bye