Question

When the cube of a non-zero number \(y\) is subtracted from \(35,\) the result is equal to the result of dividing 216 by the cube of that number \(y .\) What is the sum of all the possible values of \(y ?\) a. \(\frac{5}{2}\) b. 5 c. 6 d. 10 e. 12

Short answer

The sum of all possible values of \(y\) is 5.

Step-by-step solution

Set up the equation

Translate the problem statement into a mathematical equation. Given: When the cube of a non-zero number \(y\) is subtracted from \(35\), the result is equal to the result of dividing \(216\) by the cube of that number \(y\). Thus, the equation is: 35 - y^3 = \frac{216}{y^3}.

Eliminate the fraction

Multiply both sides of the equation by \(y^3\) to eliminate the fraction: y^3 (35 - y^3) = 216.

Simplify the equation

Distribute \(y^3\) on the left-hand side and simplify: 35y^3 - y^6 = 216 Rearrange to form a standard polynomial equation: y^6 - 35y^3 + 216 = 0.

Solve for \(z = y^3\)

Introduce a substitution \(z = y^3\) to turn the equation into a quadratic form: z^2 - 35z + 216 = 0.

Solve the quadratic equation

Solve for \(z\) using the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = -35\), and \(c = 216\). \[ z = \frac{35 \pm \sqrt{35^2 - 4 \cdot 1 \cdot 216}}{2 \cdot 1} \] \[ z = \frac{35 \pm \sqrt{1225 - 864}}{2} \] \[ z = \frac{35 \pm \sqrt{361}}{2} \] \[ z = \frac{35 \pm 19}{2} \] Thus, the solutions are \(z = 27\) and \(z = 8\).

Determine the values of \(y\)

Since \(z = y^3\), find the corresponding values of \(y\): If \(z = 27\), then \(y^3 = 27\) implies \(y = 3\). If \(z = 8\), then \(y^3 = 8\) implies \(y = 2\).

Sum all possible values of \(y\)

Finally, find the sum of all possible values of \(y\): \(\text{Sum} = 3 + 2 = 5\).

Key concepts

Cube of a Number

The cube of a number refers to raising a number to the third power. For a given number \(y\), its cube is represented as \(y^3\). This means multiplying the number by itself twice more: \(y \times y \times y = y^3\).

For example:

• If \(y = 2\), then \(2^3 = 2 \times 2 \times 2 = 8\).
• If \(y = 3\), then \(3^3 = 3 \times 3 \times 3 = 27\).

In the given problem, you need to solve an equation where a cube of a number plays a key role in both subtracting from 35 and dividing a constant.

Polynomial Equation

A polynomial equation involves variables raised to different powers with constant coefficients. They can range from simple (linear) to complex (higher-degree polynomials). The equation from our problem is an example of a polynomial equation:

When we arrived at
\[ y^6 - 35y^3 + 216 = 0 \], we had a polynomial equation of degree 6. These types of equations often have multiple solutions, as each root corresponds to a potential value of the variable.

Quadratic Formula

The quadratic formula is used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). The solutions for \(x\) can be found using the formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In the problem, substituting \(z = y^3\) converted our original polynomial to a quadratic form \(z^2 - 35z + 216 = 0\).

Here,

• \(a = 1\)
• \(b = -35\)
• \(c = 216\)

. Using the quadratic formula, we found the solutions for \(z\) which helped us find the values for \(y\).

Substitution Method

The substitution method involves replacing one variable with another expression to simplify an equation. This is particularly helpful with polynomial equations that can be transformed into quadratic equations.

In this exercise, we introduced a substitution \(z = y^3\). This substitution recast our original sixth-degree polynomial into the quadratic equation \(z^2 - 35z + 216 = 0\).

Solving the quadratic equation for \(z\) then allowed us to back-substitute to find the values of \(y\). This technique is invaluable for transforming and solving higher-degree equations in mathematics.