Question

determine the number of real solutions of the equation. then select the solution(s).
$x^3 = 27$
the equation has real solution(s)

Explanation:

Step1: Recall the definition of cube root

For the equation \(x^3 = 27\), we know that if \(a^3=b\), then \(a=\sqrt[3]{b}\). So we need to find the number whose cube is 27 and also check if there are other real solutions.

Step2: Calculate the cube root of 27

We know that \(3\times3\times3 = 3^3=27\), and for the cubic equation \(y = x^3\), the function is one - to - one (its derivative \(y^\prime=3x^2\geq0\) for all real \(x\), and it is strictly increasing since \(y^\prime = 0\) only at \(x = 0\) and the function is increasing on \((-\infty,0)\) and \((0,+\infty)\)). So the equation \(x^3=27\) has exactly 1 real solution.
To find the solution, we solve \(x=\sqrt[3]{27}\). Since \(3^3 = 27\), we have \(x = 3\).

Answer:

The equation \(x^3 = 27\) has 1 real solution, and the solution is \(x = 3\)