For this case we have a function of the form [tex]y = f (x)[/tex], where[tex]f (x) = (x-12) ^ 3-10[/tex]
To find the real zeros we must equal zero and clear the variable "x".
[tex](x-12) ^ 3-10 = 0[/tex]
We add 10 to both sides of the equation
[tex](x-12) ^ 3-10 + 10 = 10\\(x-12) ^ 3 = 10[/tex]
We apply cube root to both sides of the equation:
[tex]\sqrt[3]{(x-12)^3} = \sqrt[3] {10}\\x-12 = \sqrt[3] {10}[/tex]
We add 12 to both sides of the equation:
[tex]x-12 + 12 = \sqrt[3] {10} +12\\x = \sqrt[3] {10} +12[/tex]
Answer:
Option D
