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