"Show that x^6 - 7x^3 - 8 = 0 has a quadratic form. Then find the two real roots and the four imaginary roots of this equation."

I used synthetic division to get the real roots 2 and -1, but I can't figure out how to get the imaginary roots. I checked HotMath but all it showed me was how to get the real roots and then it practically gave me the imaginary roots without showing how to work it out. Does anyone know how to do this? Any help is GREATLY appreciated!! :D

how about this

let y = x^3
then your equation becomes
y^2 - 7y - 8 = 0
(y-8)(y+1) = 0
so y = 8 or y = -1

then x^3 = 8 or x^3 = -1
x^3 - 8 = 0
(x-2)(x^2 + 2x + 4) = 0
this will give you the real root of x=2 and 2 imaginary roots from the quadratic

or
x^3 + 1 = 0
(x+1)(x^2 - x + 1) = 0
here x = -1 plus 2 more imaginary roots

Ohhhh duh!! I can't believe I totally overlooked that. Thanks Reiny!! :D

Well, well, well! Solving equations can be quite a circus, can't it? Don't worry, my friend, Clown Bot is here to help you find those elusive imaginary roots!

To start, let's factor the equation a little differently. We have x^6 - 7x^3 - 8 = 0, right? So, can you spot anything amusing about this equation? I'll give you a moment to think...

Ah, got it? Great! Notice that if we let y = x^3, we can rewrite the equation as y^2 - 7y - 8 = 0. Now this transformed equation has a nice, quadratic form!

To find the roots of this quadratic equation, we can use the good old quadratic formula. The formula goes like this: x = (-b ± √(b^2 - 4ac)) / 2a.

For our equation y^2 - 7y - 8 = 0, we have a = 1, b = -7, and c = -8. Plugging these values into the quadratic formula, we get:

y = (-(-7) ± √((-7)^2 - 4(1)(-8))) / (2(1))

Simplifying that a little, we have:

y = (7 ± √(49 + 32)) / 2

Or:

y = (7 ± √81) / 2

Which further simplifies to:

y = (7 ± 9) / 2

So the values of y are:

y = (7 + 9) / 2 = 8

and

y = (7 - 9) / 2 = -1

Remember, y = x^3, so we substitute these values back in to find the corresponding x values:

For y = 8, we have x^3 = 8. Therefore, x = ∛8 = 2. Voila! One real root found.

For y = -1, we have x^3 = -1. Therefore, x = ∛(-1) = -1. And ta-da! Another real root discovered.

Now, as for the four imaginary roots, we need to solve x^3 = -8. By using complex numbers, we can express -8 as 8 * (cosπ + i sinπ).

So, by De Moivre's theorem, the three cube roots of -8 are:

∛(8 * (cosπ + i sinπ))

= 2 * ∛(cos(π/3 + 2πk) + i sin(π/3 + 2πk)), where k = 0, ±1

This gives us the imaginary roots:

x = 2 * [cos(π/3), cos(π/3 + 2π), cos(π/3 - 2π)]
and
x = 2 * [sin(π/3), sin(π/3 + 2π), sin(π/3 - 2π)]

And there you have it! The two real roots are 2 and -1, while the four imaginary roots involve some trigonometric fun. Remember, solving equations is a circus act, so enjoy the show and keep practicing! :D

To find the imaginary roots of the equation x^6 - 7x^3 - 8 = 0, we can use a method called substitution. Let's set a new variable y = x^3.

First, rewrite the equation using the substitution:
y^2 - 7y - 8 = 0

Now, we can solve this quadratic equation for y. We can either factor it or use the quadratic formula.

Factoring:
(y - 8)(y + 1) = 0

This gives us two possible values for y:
1) y - 8 = 0 → y = 8
2) y + 1 = 0 → y = -1

Now, let's substitute the values of y back into the original substitution equation, y = x^3, to find the corresponding values of x:

For y = 8:
x^3 = 8
Taking the cube root:
x = 2

For y = -1:
x^3 = -1
Taking the cube root:
x = -1

So, we have found the two real roots of the equation as x = 2 and x = -1.

To find the imaginary roots, we need to use complex numbers. Recall that if we have a quadratic equation of the form ax^2 + bx + c = 0, and the discriminant (b^2 - 4ac) is negative, the equation has imaginary roots.

In our case, the quadratic equation y^2 - 7y - 8 = 0 has two real roots y = 8 and y = -1. Therefore, there are no imaginary roots for the original equation x^6 - 7x^3 - 8 = 0.

Hence, the given equation only has two real roots, x = 2 and x = -1.