Question

(х-2)3 - 27 = 0 P.S. "3" - это то что в скобочках в кубе

Answer 1

смотрим по формулам кубическим-

раскрываем  скобки получаем( х^3-3*2*x^2+3*x*2^2-2^3 )- 27=0

упрощаем

x^3-6x^2+12x-19=0

 а потом просто решать как кубическое

 

Answer 2

(x-2)^3-3^3=0

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

(x-5)(x^2-x+7)=0

x=5

Answer 3

To solve the equation, we can rewrite it as follows:

(х - 2)^3 - 27 = 0

Let's simplify the equation step by step:

Expanding (х - 2)^3 using the binomial cube formula, we get:

(х - 2)^3 = (х)^3 - 3(х)^2(2) + 3(х)(2)^2 - (2)^3 = х^3 - 6х^2 + 12х - 8

Now we substitute this expression back into the original equation:

х^3 - 6х^2 + 12х - 8 - 27 = 0

Combining like terms:

х^3 - 6х^2 + 12х - 35 = 0

At this point, we need to find the values of x that satisfy this equation. Unfortunately, this equation cannot be easily factored or solved algebraically.

To find the numerical solutions, we can use numerical methods such as graphing or iteration techniques like Newton's method. These methods provide approximate solutions.

If you need a more accurate solution or have any further questions, please let me know!