Oxford University Entrance Exam Tricks probably you didn't know EXISTED

[@raghvendrasingh1289]

(x+2)^6 = 4^6
We put x+2 = 4 y
then y^6 = 1
y^3 = 1 ,-1
y = 1 , -1, w , w^2 , - w , - w^2
x= 4 y - 2 hence
x = 2 , -6 , 4 w-2 , 4 w^2 -2 , -4 w-2 , - 4 w^2 -2
x= 2 , - 6 , - 4+2√3 i , - 4-2√3 i , - 2 i√3 , 2 i√3

[@cutecat986]

let (x+2)=A : A^6=2^12 ; A=(2^12)^1/6 ; A=2^2 or A=4, as A=x+2 or 4=X+2, so X=2

[@wes9627]

x_j = 4[cos(jπ/3) + i*sin(jπ/3)]-2, j=0, 1, ..., 5

[@SBMalik-s3q]

Sorry sir, but I have a querie
Can't it be like that:
(x+2)⁶ = 2¹²
=x+2 = 6rooted 2¹²
=x+2 = 2^12/6
=x+2 = 2²
=x+2 = 4
=x =4-2
=x =2
??

[@SBMalik-s3q]

Can't it be like that:
(x+2)⁶=2^12
=x+2 = 6 rooted (2)¹²
=x+2 = (2)^12/6
=x+2 = (2)²
=x+2 = 4
=x = 4 -2
=x = 2

[@SBMalik-s3q]

Sorry sir, but I have a querie
Can't it be like that:
(x+2)⁶ = 2¹²
=x+2 = 6rooted 2¹²
=x+2 = 2^12/6
=x+2 = 2²
=x+2 = 4
=x =4-2
=x =2
??😅

[@SBMalik-s3q]

Sorry sir, but I have a querie
Can't it be like that:
(x+2)⁶ = 2¹²
=x+2 = 6rooted 2¹²
=x+2 = 2^12/6
=x+2 = 2²
=x+2 = 4
=x =4-2
=x =2
??😅

[@SBMalik-s3q]

Sorry sir, but I have a querie
Can't it be like that:
(x+2)⁶ = 2¹²
=x+2 = 6rooted 2¹²
=x+2 = 2^12/6
=x+2 = 2²
=x+2 = 4
=x =4-2
=x =2
??😅

[@SBMalik-s3q]

Sorry sir, but I have a querie
Can't it be like that:
(x+2)⁶ = 2¹²
=x+2 = 6rooted 2¹²
=x+2 = 2^12/6
=x+2 = 2²
=x+2 = 4
=x =4-2
=x =2
??😅

[@SBMalik-s3q]

Sorry sir, but I have a querie
Can't it be like that:
(x+2)⁶ = 2¹²
=x+2 = 6rooted 2¹²
=x+2 = 2^12/6
=x+2 = 2²
=x+2 = 4
=x =4-2
=x =2
??😅

[@SBMalik-s3q]

Sorry sir, but I have a querie
Can't it be like that:
(x+2)⁶ = 2¹²
=x+2 = 6rooted 2¹²
=x+2 = 2^12/6
=x+2 = 2²
=x+2 = 4
=x =4-2
=x =2
??😅

[@superacademy247]

You've found one solution. You need to find six roots!