Question

Z^6-64=0

Answer 1

it is clear that you're expected to find both the real and complex roots. I find the easiest way to find complex roots is to use the polar form of a complex number (and I'm going to replace "x" with "z" because that's the conventional notation). The polar form of a complex number is z = r • exp{iθ}, r real and r ≥ 0. exp{iθ} is a complex number with magnitude = 1 (i.e., |exp{iθ}| = 1).

z⁶ = (r • exp{iθ})⁶ = r⁶ • exp{6iθ} = 64 - I've just used the property of exponents.

Since 64 is real and has magnitude = 64, we know immediately that r = 2 (the only positive real root). We also know that exp{6iθ} = 1 = exp{2nπi} (n = integer). Thus, 6θ = 2nπ

There are six possible distinct solutions for θ: θ = 0, 2π/6, 4π/6, 6π/6, 8π/6, 10π/6

z (= x) = 2 exp{iθ}, θ = 0, 2π/6, 4π/6, 6π/6, 8π/6, 10π/6

The numbers exp{iθ} here are called the 6 roots of unity and are very easily obtained graphically. See the links below.

You can get the answers in the form Geezah gave by just plugging into

exp{iθ} = cosθ + isinθ
exp{2πi/6} = cos(2π/6) + isin(2π/6) = (1/2) + (√3/2)i, etc...

Note: Since Geezah clearly has the best math background among the other answerers so far, I suspect he may really be doing the problem my way in his head before giving you the other way – virtually everyone who uses complex numbers routinely does it the way I've shown you. It requires a little more fundamental knowledge, but it's so much easier to just multiply and divide exponents than it is to mess around with nasty factoring of large order polynomials! You really can figure out the answer in your head, once you understand the "N roots of unity" concept - see the links below.
Source(s):
http://mathworld.wolfram.com/RootofUnity …

http://en.wikipedia.org/wiki/Root_of_uni …

Answer 2

capish :)

Answer 3

if you want it a bit more towards the answer
First add 64 to both sides of the equation so you have:
x^6 = 64

Next, you will want to take the 6th root of both sides (instead of square root, do 6th root). ALTERNATELY, you can think of this as raising both sides to the 1/6th power, which is the same as taking the 6th root.

sixthroot(x^6) = sixthroot(64)

OR

(x^6)^(1/6) = 64^(1/6)

The 6th power or exponent and the 6th root on the x on the left side essentially cancel each other out.

You know how when you raise a power to a power you multiply? Like (x^2)^3 = x^6.... So the exponents did 2*3 = 6. Well, look at the previous answer's left side (x^6)^(1/6). you are raising a power to a power, so you can multiply the exponents so it's 6*(1/6) = 1. Now the exponent is 1, so you have made it a normal x!

Now you are almost done... Just need to figure out what the 6th root of 64 is, or 64^(1/6).

x = 64^(1/6)

If you had square root of 4, you would think what number multiplied twice gives you 4? Like 2*2 = 4, so the square root of 4 (really 4^1/2) = 2. Must do this for the 6th root of 64.

2*2 = 4
2*2*2 = 8
2*2*2*2 = 16
2*2*2*2*2 = 32
2*2*2*2*2*2 = 64

Well, what do ya know? When you do 2^6, you get 64! This means that the 6th root of 64 = 2.

So, after much ado, your final answer is:

x = 2
Source(s):
me brain