What is the solution to the equation?
$\sqrt[3]{x-1}=-4$
The problem asks us to solve the equation \(\sqrt[3]{x-1} = -4\) for \(x\). The variable \(x\) is a real number. The cube root function \(\sqrt[3]{u}\) is defined for all real numbers and is an injective (one-to-one) function. To solve for \(x\), we need to isolate it by eliminating the cube root. This can be achieved by raising both sides of the equation to the power of three.
To eliminate the cube root on the left side of the equation \(\sqrt[3]{x-1} = -4\), we cube both sides:
\[(\sqrt[3]{x-1})^3 = (-4)^3\]
The cube of a cube root simplifies the expression:
\[x - 1 = (-4) \times (-4) \times (-4)\]
Calculating the right side, \((-4) \times (-4) = 16\), and \(16 \times (-4) = -64\).
So the equation becomes:
\[x - 1 = -64\]
Now we have a simple linear equation:
\[x - 1 = -64\]
To solve for \(x\), we add 1 to both sides:
\[x = -64 + 1\]
\[x = -63\]
Comparing this result to the given options, we see that \(-63\) is one of the choices provided. Therefore, the solution to the equation \(\sqrt[3]{x-1} = -4\) is \(-63\).
In real-world scenarios, cube root equations often arise when dealing with volume calculations. For instance, if you are designing a storage box in the shape of a cube and you know the volume is a negative value relative to a baseline, you might encounter similar structures. Although physical dimensions like volume cannot be negative, such equations are essential in fields like physics or economics where values can represent changes relative to a reference point, such as "three units below a zero-baseline level." Mastering these algebraic manipulations allows for solving complex growth models or rate-of-change problems across various scientific disciplines.