whats wrong with this?

Question

Sqrt=square root and i=square root of -1
(-1+Sqrt(3)i)^3= 8 this line is right
-1+Sqrt(3)i = 2 i took the cube root of each side
Sqrt(3)i = 3 added 1 to each side
i = Sqrt(3) divided both sides by the Sqrt(3)
-1 = 3 square each side
obviosly the last line is not true but then where is the mistake?

Answer 1

The problem lies in step 2.
Consider the equation
x^3 = 8
This equation has 3 distinct (unequal) solutions (one is real and the other two are imaginary)
The real solution is the easiest to find. It is two.
Another solution is the one from the problem above.
-1 + sqrt(3)i
The third solution is the conjugate of -1 + sqrt(3)i
It is -1 - sqrt(3)i

A similar problem is
(-2)^2 = 4
Take the square root of each side
-2 = 2
This is not true.
When you are taking the cubed root of 8, you are only considering the real value. In this case, you have to consider the imaginary values as well.

Answer 2

This is the same as the ff: only in a more complex scale:
-8 = -8
4 - 12 = 16 - 24
4 - (6 + 6) = 16 - (12 + 12)
4 - (6 + 6) + 9 = 16 - (12 + 12) + 9
(2 - 3)² = (4 - 3)²
2 - 3 = 4 - 3
2 = 4

Obviously the wrong part was when I got sqrt of both sides. It is clear that the squares are equal, but that square has 2 DIFFERENT roots. So you cannot get square root of both sides.

here i only used sqrt. and you used cube root. In your example you cannot take
(1 ± sqrt3 i)³ = 8
Cube root of both sides because it is equivalent to
(1 ± sqrt3 i)³ = 2³

Here, 1 ± sqrt3 i and 2 are all 3 DIFFERENT roots of 8. You cannot equate them, for the same reason as sqrt4 and - sqrt4 are not equal.

^_^

Answer 3

Your first line is correct. The mistake lies in your assumption that if x^3=y^3 then x=y. This is not true - just look at your second line. Does -1+i√3 look like it's equal to 2? Perhaps this mistake will be easier to see if we use square roots instead of cube.

(-1)²=1²
-1=1
0=2 ← this is wrong.

Generally speaking, in the complex numbers, an nth root will have n distinct solutions, and you cannot assume that any two of these solutions will actually be the same. If your solution to a problem involves taking nth roots, the most you can say is that |x|=|y|, and that if you represent the numbers in polar from, the argument of x will be equal to the argument of y ±2πk/n, where n is the degree of the root and k is some integer. What you cannot do, as explained earlier, is assume that x and y will actually be equal - because, in the general case, they aren't.

Answer 4

although x^3 and x^(1/3) are 1-1 and invertible on the real number line, they are not so for the complex numbers. Therefore you cannot take the cube root in the second step and assume that two cubed roots are the same, because as you have shown, they aren't.

If you are dealing with real values only, then you can say "a^3=b^3 if and only if a=b," but this is not true when you extend the "field" to the complex numbers.

The fundamental theorem of algebra states that for any polynomial with complex coefficients, the roots of the polynomial are contained in the complex numbers. Since x^3-8=(x-2)(x^2+2x+4) has three distinct roots (this can be seen by noticing that the discriminant of x^2+2x+4 2^2-4(1)(4)=-12<0). Therefore by the FTA, it has three distinct solutions in C, and is thus not invertible.

Again, x^3-8 IS invertible in the reals, but NOT in the complex numbers.

I'm sure that you knew/know that x^2+b is not invertible in the reals, and that x^3+b is, and just thought that it worked, but I hope now you know that for any polynomial of degree >1, it will not be invertible over C (or more specifically over any field that contains more than one of its roots).

I hope that helps, if some of the stuff is too confusing, don't think too much about the hard definitions I gave, just wanted to try and be a little thorough.

Answer 5

I do not believe the first line is an equality.

In that case, none of the other lines are equalities.

The term with i in it is raised to the power of 3. That means there will still be a term with i in it. That means it cannot equal 8.
In other words, if you expand (-1_sqrt(3)i)^3, you still have a complex number.

For the initial equation to be an equality sqrt(3)i-1=2 would have to be true, and
sqrt(3)i=3 which is not so

Answer 6

if you divide both sides by the square root of 3 then you will have a square root in the denominator and you can't have that. You have to then multiply the numerator by opposite the denominator and the denominator by the square root of 3.

Sqrt(3)i = 3
i = 3 / sqrt(3)
i = 3* -[sqrt(3)] / sqrt(3)*sqrt(3)
i = -3sqrt(3) / sqrt(9)
i = -3sqrt(3) / 3
i = -1
-1 = -1

Answer 7

You may want to start by multiplying (-1+sqrt(3)i) X (-1+sqrt(3)i)
in order to get (-1+sqrt(3)i)^2. Then multiply that value by
(-1+sqrt(3)i) in order to obtain (-1+sqrt(3)i)^3. Be careful, so as not to make any arithmetic mistakes. Hope this is useful to you.

Answer 8

im confused here. I follow the math, but it looks as if you are trying to solve for something. Are you trying to show that both sides are equal? I would look more closely at the 'Sqrt(3)i'.

Answer 9

you are wrong when u equalise -1+sqrt(3)i to 2
because the former is an imaginary number... and 2 is a real number.

the reason is explained by mathgirl.. she know her thing very well

Answer 10

I don't understand how (-1+sqrt(3)i)^3=8

or simpler, sqrt(3)i=3

the sqrt of 3 * i cannot be 3 ----