If only it were as easy as 1, 2, 3

Some good solution has been posted,yet i am writing this because this is different(i think so)

$z\quad can't\quad be\quad 0.\\ \\ just\quad multiply\quad both\quad side\quad by\quad z+\cfrac { 1 }{ z } \\ \\ \left( { z }^{ 2 }+\cfrac { 1 }{ { z }^{ 2 } } \right) +2={ z }^{ 3 }+\cfrac { 1 }{ { z }^{ 3 } } +\left( z+\cfrac { 1 }{ z } \right) \\ \\ { z }^{ 3 }+\cfrac { 1 }{ { z }^{ 3 } } =2$

Before I begin my solution, let me explain what's so special about $z+\frac{1}{z}$ .

First, consider the equation

$z+\frac{1}{z}=2\cos \theta$

Solve for $z$ , we get

$z=\cos\theta + i\sin \theta$

Looks familiar? Yes, it is the well-known DeMoivre's Theorem !

Hence, for any integer $n$ , we have

$z^n+\frac{1}{z^n}=2\cos {n\theta}$

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Using the method stated above,

$z+\frac{1}{z}=z^2+\frac{1}{z^2}$

$2\cos \theta = 2\cos{2\theta}$

Solve the equation we will get

$\theta=120^\circ$

*Note that the solution $\theta=0^\circ$ is not suitable in this case.

Substitute to the expression

$z^3+\frac{1}{z^3}=2\cos{3(120^\circ)}$

We get the desired answer $2$ .

$Rearranging\ the\ given\ terms\ we\ get:\\ z-z^{2}=\dfrac{1}{z^{2}}-\dfrac{1}{z}\\ \Rightarrow\ z*(1-z)=\dfrac{z*(1-z)}{z^{3}}\\ We\ can\ cancel\ (z)\ and\ (1-z)\ as\ they\ can't\ be\ equal\ to\ zero\ as\ z\ is\ a\ \\complex\ number.\\ \Rightarrow\ z^{3}=1.\\ \Rightarrow\ \dfrac{1}{z^{3}}=1.\\ \Rightarrow\ z^{3}+\dfrac{1}{z^{3}}=1+1=\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{2}}}}}}}}}}}}}}}}}}.$

$z=\omega$ satisfies the given condition.

So, $z^3+\frac{1}{z^3} =\omega^3+\frac{1}{\omega^3}=1+1=\boxed{2}$

It was so easy, u need a hawk eye to do this, just look at the sentence "If only it was as easy as z=1

So:

$(1)^3+\frac{1}{(1)^3}=1+1=\boxed{2}$

Begin by factoring the equation to $0=z(z-1)-\frac{1}{z^2}(z-1)$

$0=(z-\frac{1}{z^2})(z-1)$

we can disregard the (z-1) because the question states that z is complex. Thus we are looking at $0=(z-\dfrac{1}{z^2})$

Multiplying both sides by $z^{\frac{1}{2}}$ we get $0=(z^{\frac{3}{2}}-\dfrac{1}{z^{\frac{3}{2}}})$

squaring both sides we get $0=z^3+\dfrac{1}{z^3}-2$

$\therefore \boxed{2}=z^3+\dfrac{1}{z^3}$

• $z+\frac{1}{z}=z^2+\frac{1}{z^2} \longrightarrow 1$

Multiply Eq. $1$ by $z$ :

• $z^2+1=z^3+\frac{1}{z} \longrightarrow 2$

Divide Eq. $1$ by $z$ :

• $1+\frac{1}{z^2}=z+\frac{1}{z^3} \longrightarrow 3$

Adding 2 & 3 :

• $(z^2+\frac{1}{z^2})+2=(z+\frac{1}{z})+(z^3+\frac{1}{z^3})$ $\Rightarrow (z^3+\frac{1}{z^3})=\boxed{2}$

Let $z+\frac{1}{z} = z^2+\frac{1}{z^2} = x$

We know that $x = x^2-2 \implies x^2-x-2 = 0 \implies x = \frac{1\pm3}{2}$

Since, according to the problem, $x<0$ , we will take the negative root: $x = -1$

Factorizing $z^3+\frac{1}{z^3}$ (sum of cubes):

$z^3+\frac{1}{z^3} = \left(z+\frac{1}{z}\right)\left(z^2-1+\frac{1}{z^2}\right) = x(x-1) = -(-2) = \boxed{2}$

rearrange the equation to obtain: Z^2 - Z = (1/Z) - (1/Z^2). The right hand side is equal to (Z^{2}-Z)/Z^3. Therefore, 1=1/z^3 and 1=Z^3. It follows that Z^3 + 1/Z^3 =2

let, z=1 then, 1^3+1/1^3 =1+1 =2

Let $z+\frac{1}{z}=a$ Then,after adding $2$ to both sides of the vequation,we get: $a+2=a^2\rightarrow\;a^2-a-2=0$ Solving,we get $a=2$ ,So: $z+\frac{1}{z}=2\rightarrow\;z^2-2z-1=0$ Solving this,we get $z=1$ so $z^3+\frac{1}{z^3}=1^3+\frac{1}{1^3}=\boxed{2}$

let k = z + 1/z = z² + 1/z²

(z + 1/z)² = k²

z² + 2 + 1/z² = k²

rearranging:

(z² + 1/z² )+ 2 = k²

k + 2 = k²

k² - k = 2

next;

(z + 1/z)(z² + 1/z²) = z³ + 1/z + z + 1/z³ = (k)(k)

rearranging:

k² = (z³ + 1/z³) + (z + 1/z)

k² = (z³ + 1/z³) + k

k² - k = (z³ + 1/z³)

2 = (z³ + 1/z³)

ANSWER is "2".

Z is a cube root of unity.. And z≠1.z=w,w^2.. Place and solve.solution =2..

It is given that: $z+\dfrac {1}{z} = z^2 + \dfrac {1}{z^2} < 0$

Consider $\left( z+\dfrac {1}{z} \right) ^2 = z^2+ 2 + \dfrac {1}{z^2} = z+\dfrac {1}{z} + 2$

$\Rightarrow \left( z+\dfrac {1}{z} \right) ^2 - \left( z+\dfrac {1}{z} \right) - 2 = 0$

$\Rightarrow \left( z+\dfrac {1}{z} \right) = \dfrac {1 \pm \sqrt{1+8}}{2} = \dfrac {1 \pm 3}{2} = -1 < 0$

Now $\left( z+\dfrac {1}{z} \right) \left( z^2+\dfrac {1}{z^2} \right) = z^3 + z+\dfrac {1}{z} + \dfrac {1}{z^3}$

$\Rightarrow (-1)(-1) = z^3 + \dfrac {1}{z^3} - 1$

$\Rightarrow z^3 + \dfrac {1}{z^3} = (-1)(-1) +1 = \boxed{2}$