Factor the expression on the lift side of each equation. Then solce the equation: 8xcubed-27=0... also please show work so I understand how you got the answer.
I am sure that in your course you are dealing with a factoring pattern called the sum/difference of cubes.
Look in your text and you will find something like
A^3 - B^3 = (A-B)(A^2 + AB + B^2)
Your equation fits that pattern, let me know how you answered it.
We can't take our text book home. Do you think you could explain this to me?
Sorry, I was letting someone use my computer, and their name showed up on the last post, but it was me that posted it.
To factor the expression and solve the equation 8x^3 - 27 = 0, we can use the difference of cubes formula, which states that a^3 - b^3 can be factored as (a - b)(a^2 + ab + b^2). In this case, a is 2x and b is 3.
Step 1: Write the given equation: 8x^3 - 27 = 0
Step 2: Apply the difference of cubes formula:
(a - b)(a^2 + ab + b^2) = (2x - 3)(4x^2 + 6x + 9)
Step 3: Set each factor equal to zero and solve for x:
2x - 3 = 0 (Equation 1)
4x^2 + 6x + 9 = 0 (Equation 2)
Let's solve each equation separately:
Equation 1: 2x - 3 = 0
Adding 3 to both sides:
2x = 3
Divide both sides by 2:
x = 3/2 or x = 1.5
Equation 2: 4x^2 + 6x + 9 = 0
This equation cannot be factored further, so we need to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 4, b = 6, and c = 9. Substituting these values into the formula:
x = (-6 ± √(6^2 - 4 * 4 * 9)) / (2 * 4)
Simplifying:
x = (-6 ± √(36 - 144)) / 8
x = (-6 ± √(-108)) / 8
Since we have a negative value under the square root, the equation has no real solutions, only imaginary solutions. Therefore, Equations 2 does not have any real solutions.
In conclusion, the solutions to the equation 8x^3 - 27 = 0 are:
x = 3/2 or x = 1.5