it is very easy and takes 5 minutes let a= (x-2)/3 a⁶=1 a=1^(1/6) the sixth roots of unity are well known =exp(i(2n+1)π)/3 = +-1 and +(1/2)+-i(ζ3/2) -(1/2)+-i(ζ3/2) x=3a+2 for all 6 values of a
Note that 729=3⁶ then (x-2)⁶=3⁶. If x is real the x-2=±3 --> x=2±3 To find all six roots are as follows: (x-2)⁶-3⁶=0 [(x-2)³+3³][(x-2)³-3³]=0 • (x-2)³+3³=0 [(x-2)+3][(x-2)²-3(x-2)+3²]=0 --> x=-1 (x-2)²-3(x-2)+3²=0 (x-2)=½[3±3isqrt(3)] x=2+½[3±3isqrt(3)] =½[7±3isqrt(3)] • (x-2)³-3³=0 [(x-2)-3][(x-2)²+3(x-2)+3²]=0 --> x=5 (x-2)²+3(x-2)+3²=0 (x-2)=½[-3±3isqrt(3)] x=2+½[-3±3isqrt(3)] =½>1±3isqrt(3)]
This doesnt need any trick. There are 6 solution of 6th root distributed evenly along the circle of radius 3. All you need to know is cosinus of 60 degree which is 1/2
DeMoivre's Theorem is very useful here .
it is very easy and takes 5 minutes
let a= (x-2)/3
a⁶=1
a=1^(1/6)
the sixth roots of unity are well known =exp(i(2n+1)π)/3
=
+-1
and
+(1/2)+-i(ζ3/2)
-(1/2)+-i(ζ3/2)
x=3a+2 for all 6 values of a
It's nice when someone has a good understanding of complex numbers and doesn't waste time on long transformations.
5,-1,11,-7
29,-25
5
Note that 729=3⁶ then (x-2)⁶=3⁶. If x is real the x-2=±3 --> x=2±3
To find all six roots are as follows:
(x-2)⁶-3⁶=0
[(x-2)³+3³][(x-2)³-3³]=0
• (x-2)³+3³=0
[(x-2)+3][(x-2)²-3(x-2)+3²]=0
--> x=-1
(x-2)²-3(x-2)+3²=0
(x-2)=½[3±3isqrt(3)]
x=2+½[3±3isqrt(3)]
=½[7±3isqrt(3)]
• (x-2)³-3³=0
[(x-2)-3][(x-2)²+3(x-2)+3²]=0
--> x=5
(x-2)²+3(x-2)+3²=0
(x-2)=½[-3±3isqrt(3)]
x=2+½[-3±3isqrt(3)]
=½>1±3isqrt(3)]
X=5
X=-1 and x=5
729 = 9(81) = 9(9)(9) = 3^6
(x - 2)^6 = 3^6
y = x - 2
(y^3 + 3^3)(y^3 - 3^3) = 0
(y + 3)(y^2 - 3y + 3^2)(y - 3)(y^2 + 3y + 3^2) = 0
y = +/- 3, (3 +/- 3i√3)/2, (-3 +/- 3i√3)/2
x = y + 2 = -1, 5, (7 +/- 3i√3)/2, (1 +/- 3i√3)/2
(x -2)^6 = 729, x = ?
*. 729 = 3x3x3x3x3 = 3^6
(x-2)^6 = 3^6
x - 2 = 3.
**. x = 3 + 2 = 5 ./.
Excellent job 👏
2 real roots, 4 complex roots
(x-2)⁶=729
Let y=x-2
y⁶=729
y⁶-729=0
(y³+27)(y³-27)=0
(y+3)(y-3)(y²-3y+9)(y²+3y+9)=0
y²-3y+9=0
4y²-12y+36=0
4y²-12y+9=-27
(2y-3)²=-27
|2y-3|=3i√3
2y-3=±3i√3
2(x-2)-3=±3i√3
2x-4-3=±3i√3
2x-7=±3i√3
2x=7±3i√3
x=½(7±3i√3) ❤❤
y²+3y+9=0
4y²+12y+36=0
4y²+12y+9=-27
(2y+3)²=-27
|2y+3|=3i√3
2y+3=±3i√3
2(x-2)+3=±3i√3
2x-4+3=±3i√3
2x-1=±3i√3
2x=1±3i√3
x=½(1±3i√3) ❤❤
y+3=0
y=-3
x-2=-3
x=-1 ❤
y-3=0
y=3
x-2=3
x=5 ❤
Excellent
This doesnt need any trick. There are 6 solution of 6th root distributed evenly along the circle of radius 3. All you need to know is cosinus of 60 degree which is 1/2
I'd have thought everyone knew 3 to the 6th was 729,, doesn't seem a good question, you should specify the real solutions are required
5, it took 30 seconds....but oh yeah you want the 5 imaginary solutions I forgot
Minus 1 is hardly a imaginary solution