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I just asked myself. if you have:

$$x^3 -8 = 0$$ and trying to solve it, you do

$$x^3 = 8$$

well, the first real root will be $2$ because $2^3 = 8$

but it is only the first, what about second and third?

ab123
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d.sci
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  • They will be complex roots. You can show that there is a contradiction if you assume that the second and third roots are real. – Sarvesh Ravichandran Iyer Nov 05 '17 at 11:09
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    Use the factorisation: $x^3 - 8 = (x-2)(x^2 + 2x + 4)$ – jonsno Nov 05 '17 at 11:11
  • Use De Moivre formula (https://en.wikipedia.org/wiki/De_Moivre%27s_formula, paragraph on roots of complex numbers) to get all the solutions in a row. – Gibbs Nov 05 '17 at 11:13
  • Thats better way. Also note that $x^3$ is strictly increasing for $x > 0$ so $y =x^3$ and $y = 8$ cannot have more than one point of intersection. – jonsno Nov 05 '17 at 11:16

5 Answers5

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HINT: use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Dr. Sonnhard Graubner
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We know $x = 2$ is one root, so $f (x) $ has a factor of $(x-2) $. Simplifying, we have, $$x^3 = 8$$ $$\implies x^3-8=0$$ $$\implies (x-2)(x^2+2x+4) = 0$$

Now you can solve the quadratic by the quadratic formula. Its roots are complex.

1

If $x\in\mathbb C$ is such that $x^3=8$, then $|x|^3=|x^3|=8$ and therefore $|x|=2$. So, write $x$ as $2(\cos\theta +i\sin\theta)$ ($\theta\in[0,2\pi)$). Then $x^3=8\bigl(\cos(3\theta)+i\sin(3\theta)\bigr)$. When is this equal to $8$? That's when $\theta=0$, $\theta=\frac{2\pi}3$ or $\theta=\frac{4\pi}3$.

José Carlos Santos
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$$x^3-8=0$$

One of the root is $2$. The complex root comes in pair when the coefficient is real.

From Vieta's formula,

$$2+(a+ib)+(a-ib)=0 \implies 2a+2=0 \implies a=-1$$

$$2(a+ib)(a-ib)=8 \implies a^2+b^2=4$$

You just have to solve for $b$.

Siong Thye Goh
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0

Let $x=u+iv$. Then cubing and equating real/imaginary parts,

$$u^3-3uv^2=8,\\3u^2v-v^3=0.$$

The second equation tells us that $v=0$ or $v=\pm\sqrt3u$. And by the first,

  • $v=0\to x=2$,

  • $v=\sqrt u\to -8u^3=8,x=-1-i\sqrt3$,

  • $v=-\sqrt u\to -8u^3=8,x=-1+i\sqrt3$.

(Note that only the real $u,v$ solutions of these equations can be used and there is no circular argument.)