SOLUTION: x^6+7x^3-8=0 , how do you solve for x, step by step?

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, think of this as a quadratic , where.

so, factorise to give (y+8)(y-1)=0, so then we say that either y+8=0 OR y-1=0.

Hence y=-8 or y=+1

So, OR , which means that the possible answers are:

x=-2 or x=+1

jon

You can put this solution on YOUR website!
I had started an answer on this problem but was called away from my computer before I could finish and post it.
Here's another approach to solving this. First, recognise that, since this is 6th order polynomial, you can expect to find 6 roots.
1) Factor the equation: (x^3 + 8)(x^3 - 1) = 0
Apply the zero products principle: (x^3 + 8) = 0 or (x^3 - 1) = 0
2) Notice that you now have the product of the sum of two cubes (x^3 + 2^3) and the difference of two cubes (x^3 - 1^3)
Factor the sum of two cubes: x^3 + 8 = (x + 2)(x^2 - 2x + 4) = 0
So we have: (x + 2)(x^2 - 2x + 4) = 0;
x + 2 = 0; x = -2 or
x^2 - 2x + 4 = 0; This doesn't factor so use the quadratic formula:

x = 1 + i or x = 1 - i
Factor the difference of two cubes: x^3 - 1 = (x - 1)(x^2 + x + 1) = 0
So we have: (x - 1)(x^2 + x + 1) = 0
x - 1 = 0; x = 1 or
x^2 + x + 1 = 0 This doesn't factor so use the quadratic formula:

x = -0.5 + 0.5i or x = -0.5 - 0.5i
The 6 roots are:
x = 1
x = -2
x = 1 + i
x = 1 - i
x = -0.5 + 0.5i
x = -0.5 - 0.5i