To solve the equation 125x^3 - 27 = 0, we first rewrite it as a difference of cubes and then use the quadratic formula to solve for x. The solutions are x = 3/5 and x = (-15 ± sqrt(-675)) / (50).
Step-by-step explanation:
Solution:
To solve the equation 125x^3 - 27 = 0, we first need to rewrite it as a difference of cubes:
(5x)^3 - (3)^3 = 0
Using the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2), we can rewrite the equation as:
(5x - 3)(25x^2 + 15x + 9) = 0
Setting each factor to zero, we have:
5x - 3 = 0 --> 5x = 3 --> x = 3/5
25x^2 + 15x + 9 = 0
Since the quadratic equation does not factor easily, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For this equation, a = 25, b = 15, and c = 9:
x = (-15 ± sqrt(15^2 - 4(25)(9))) / (2(25))
x = (-15 ± sqrt(225 - 900)) / (50)
x = (-15 ± sqrt(-675)) / (50)
This results in imaginary solutions for x. Therefore, the solutions to the equation 125x^3 - 27 = 0 are x = 3/5 and x = (-15 ± sqrt(-675)) / (50).