# Write cube root of3,square root of3,and the fourth root of6 in ascending order?

SHAIK AASIF AHAMED
9 years ago
Hello student,
Given (3)1/3,(3)1/2,(6)1/4
Equalizing powers by multiplying and dividing by 12 we get
=(3)(12)/(3*12),(3)12/(2*12),(6)(12)/(4*12)
=(34)1/12,(36)1/12,(63)1/12
=(81)1/12,(729)1/12,(216)1/12
As in ascending order 81,216,729
So ascending order is(3)1/3,(6)1/4,(3)1/2
Sourabh Singh IIT Patna
9 years ago
LHS=cos6A-sin6A= (cos2A)3-(sin2A)3
=(cos2A-sin2A)(cos4A+sin4A+cos2Asin2A)
[sincea3-b3=(a-b)(a2+b2+ab)]
=cos2A((cos2A+sin2A)2-cos2Asin2A
[sincecos2θ=cos2θ-sin2θ]
=cos2A.(1-1/4(2sinAcosA)2=cos2A.(1-1/4.sin22A )
[sincesin2A=2sinAcosA]
= RHS
Nilanjan Kundu
22 Points
9 years ago
First we would take the LCM of the th roots.LCM will be 12. Then we will write cube root of three like (3^4)^1/12, square root of 3 like (3^6)^1/12 and fourth root of 6 like (6^3)^1/12. THis would be
(81)^1/12, (729)^1/12 and (216)^1/12. Therefore, this in ascending order would be (81)^1/12
Aryan Vyas
37 Points
9 years ago
In such questions,we have to take the LCM of the powers first.In this case the LCM of 3,2 and 4 =12.Then we have to convert the powers to 12 and the no.which is used for this purpose has to be should be raised inside the radical,i.e.
4×3 root 34 ,6×2 root 36 , 3×4 root 63
=12 root 81, =12 root 729, =12 root 216
Now it becomes easier to rearrange as the power of the radicals are same.
Now,rearrange as we do for normal nos.
=>12 root 81
=Cube root 3