Question

0= 27x^3 + 125 Please help quickly

Answer 1

Answer:

-5/3 = x

x = 5/6 ± 5 sqrt(3)i /(6)

Step-by-step explanation:

0= 27x^3 + 125

Subtract 125 from each side

-125 = 27 x^3

Divide each side by 27

-125/27 = x^3

Take the cube root of each side

(-125/27 ) ^ (1/3) = x^3 ^ (1/3)

-5/3 = x

This is the only real answer

Let me know if you want the complex solutions

0= 27x^3 + 125 = ( 3x) ^3 + 5^3

a^3 + b^3 = (a+b) • (a^2-ab+b^2)

             0   = ( 3x+5) ( 9x^2 - 15x +25)

Using the zero product property

3x+5 =0      9x^2 - 15x +25 =0

3x = -5         9x^2 -15x = -25

x = -5/3       9(x^2 - 5/3x) = -25

Completing the square

                    (x^2 - 5/3x) = -25/9

                     x^2 - 5/3x +25/36 = -25/9 + 25/36

                    ( x-5/6) ^2 = -25/12

                    Take the square root of each side

                    sqrt( x-5/6) ^2 =± sqrt(-25/12)

                       ( x-5/6)  =± 5i/(2 sqrt(3))

                   Add 5/6

                       x = 5/6 ± 5i/(2 sqrt(3))

We cannot leave a sqrt in the denominator so multiply by sqrt(3)/sqrt(3)

                      x = 5/6 ± 5 sqrt(3)i /(2(3))

                     x = 5/6 ± 5 sqrt(3)i /(6)