Can someone help me with these three math problems?

Question

in these i half to find the imaginary solutions of each equation by factoring

8x^3 - 27=0

x^3 + 64=0

2x^3 + 54 = 0

thanks in advance!

Answer 1

1) (8x^3-27) = (2x)^3-3^3

We know the formula a^3-b^3 = (a-b)(a^2+ab+b^2)

(2x)^3-3^3 = (2x-3)[(2x)^2+(2x)(3)+3^2]

 (2x-3)(4x^2+6x+9) = 0

2x-3 = 0 and 4x^2+6x+9 = 0

2x = 3

x =  3/2

Now factorize 4x^2+6x+9

Compare it to quadratic form ax^2+bx+c = 0

a = 4, b = 6, c = 9

x = [-b±√b^2-4ac]/2a

x = [-6±√(36-144)]/8

x = [-6±√-108]/8

x = -6±√-27*4]/8

x = [-6±2√27i]/8

x = 2[-3±√27i]/8

x = [-3±√27i]/4

imaginary solutions of 4x^2+6x+9 is x = [-3±√27i]/4

Answer 2

(2).

The equation is x³ + 64 = 0.

x³ + 4³ = 0

Apply Sum of Two Cubes : u³ + v³ = (u + v)(u² - uv + v²).

(x + 4)(x² - 4x + 16) = 0.

Apply Zero Product Property : If ab = 0, then either a = 0, b = 0, or both a and b equal zero.

x + 4 = 0 and x² - 4x + 16 = 0.

x = - 4 and x² - 4x + 16 = 0.

Compare the equation x² - 4x + 16 = 0 with the standard form of quadratic equation ax² + bx + c = 0.

a = 1, b = - 4 and c = 16.

Quadratic Formula : x = [- b ± √(b² - 4ac)]/2a.

x = [-(-4) ± √{(-4)² - 4(1)(16)}]/2(1)

x = [4 ± √(16 - 64)]/2

x = [4 ± √(-48)]/2

x = [4 ± √(i²*16*3)]/2

x = 2 ± 2i√3.

The solutions are -4, 2 ± 2i√3.

Answer 3

(3).

The equation is 2x³ + 54 = 0.

x³ + 27 = 0

x³ + 3³ = 0

Apply Sum of Two Cubes : u³ + v³ = (u + v)(u² - uv + v²).

(x + 3)(x² - 3x + 9) = 0.

Apply Zero Product Property : If ab = 0, then either a = 0, b = 0, or both a and b equal zero.

x + 3 = 0 and x² - 3x + 9 = 0.

x = - 3 and x² - 3x + 9 = 0.

Compare the equation x² - 3x + 9 = 0 with the standard form of quadratic equation ax² + bx + c = 0.

a = 1, b = - 3 and c = 9.

Quadratic Formula : x = [- b ± √(b² - 4ac)]/2a.

x = [-(-3) ± √{(-3)² - 4(1)(9)}]/2(1)

x = [3 ± √(9 - 36)]/2

x = [3 ± √(-27)]/2

x = [3 ± √(i²*9*3)]/2

x = (3 ± 3i√3)/2.

The solutions are -3, 3/2 ± (3i√3)/2.