What Value Of $c$ Makes The Equation True? Assume $x \ \textgreater \ 0$ And $ Y \textgreater 0 Y \ \textgreater \ 0 Y \textgreater 0 [/tex]. X 3 C Y 4 3 = X 4 Y ( Y 3 ) \sqrt[3]{\frac{x^3}{c Y^4}} = \frac{x}{4 Y(\sqrt[3]{y})} 3 C Y 4 X 3 ​ ​ = 4 Y ( 3 Y ​ ) X ​ A. C = 12 C = 12 C = 12 B. $c =

$$

Introduction

In mathematics, equations involving variables and constants are a fundamental concept. Solving these equations requires a deep understanding of algebraic manipulations and properties of mathematical operations. In this article, we will explore a specific equation involving a variable $c$ and solve for its value. The given equation is $\sqrt[3]{\frac{x^3}{c y^4}} = \frac{x}{4 y(\sqrt[3]{y})}$, where $x > 0$ and $y > 0$. Our goal is to find the value of $c$ that makes this equation true.

Understanding the Equation

The given equation involves a cube root and fractions. To simplify the equation, we can start by isolating the cube root term. We can do this by raising both sides of the equation to the power of 3, which will eliminate the cube root.

$\left(\sqrt[3]{\frac{x^3}{c y^4}}\right)^3 = \left(\frac{x}{4 y(\sqrt[3]{y})}\right)^3$

Simplifying the Equation

When we raise both sides of the equation to the power of 3, we get:

$\frac{x^3}{c y^4} = \frac{x^3}{64 y^3 (\sqrt[3]{y})^3}$

Canceling Out Common Terms

We can simplify the equation further by canceling out common terms. Notice that $x^3$ appears on both sides of the equation, so we can cancel it out.

$\frac{1}{c y^4} = \frac{1}{64 y^3 (\sqrt[3]{y})^3}$

Simplifying the Right-Hand Side

We can simplify the right-hand side of the equation by evaluating the expression $(\sqrt[3]{y})^3$. Since $(\sqrt[3]{y})^3 = y$, we can substitute this value into the equation.

$\frac{1}{c y^4} = \frac{1}{64 y^3 y}$

Canceling Out Common Terms Again

We can simplify the equation further by canceling out common terms. Notice that $y^3 y$ appears on both sides of the equation, so we can cancel it out.

$\frac{1}{c y^4} = \frac{1}{64 y^4}$

Solving for $c$

Now that we have simplified the equation, we can solve for $c$. We can do this by cross-multiplying and then isolating $c$.

$\frac{1}{c y^4} = \frac{1}{64 y^4}$

Cross-multiplying:

$64 y^4 = c y^4$

Dividing both sides by $y^4$:

$64 = c$

Conclusion

In this article, we explored a specific equation involving a variable $c$ and solved for its value. We started by isolating the cube root term and then simplified the equation by canceling out common terms. Finally, we solved for $c$ by cross-multiplying and isolating the variable. The value of $c$ that makes the equation true is $c = 64$.

Discussion

The value of $c$ that makes the equation true is $c = 64$. This is the correct answer, and it can be verified by plugging it back into the original equation.

Alternative Solutions

There are alternative solutions to this problem, and they involve different values of $c$. However, these solutions are not correct, and they can be verified by plugging them back into the original equation.

Final Answer

The final answer is $c = 64$.

$$

Introduction

In our previous article, we explored a specific equation involving a variable $c$ and solved for its value. The equation was $\sqrt[3]{\frac{x^3}{c y^4}} = \frac{x}{4 y(\sqrt[3]{y})}$, where $x > 0$ and $y > 0$. We found that the value of $c$ that makes the equation true is $c = 64$. In this article, we will answer some frequently asked questions about the equation and its solution.

Q: What is the significance of the cube root in the equation?

A: The cube root in the equation is used to simplify the expression and make it easier to solve. By raising both sides of the equation to the power of 3, we can eliminate the cube root and simplify the equation.

Q: Why do we need to assume that $x > 0$ and $y > 0$?

A: We need to assume that $x > 0$ and $y > 0$ because the cube root of a negative number is undefined. If $x$ or $y$ were negative, the cube root would be undefined, and the equation would not make sense.

Q: Can we solve the equation for $x$ or $y$ instead of $c$?

A: Yes, we can solve the equation for $x$ or $y$ instead of $c$. However, the solution would be more complex and would involve more steps. Solving for $c$ is a simpler and more straightforward approach.

Q: What if we have a different equation with a variable $c$? Can we still use the same method to solve it?

A: Yes, we can still use the same method to solve a different equation with a variable $c$. However, we need to make sure that the equation is similar to the original equation and that the same steps can be applied.

Q: Can we use this method to solve equations with other types of roots, such as square roots or fourth roots?

A: Yes, we can use this method to solve equations with other types of roots, such as square roots or fourth roots. However, we need to be careful and make sure that we are using the correct steps and formulas for each type of root.

Q: What if we have a system of equations with multiple variables, including $c$? Can we still use this method to solve it?

A: Yes, we can still use this method to solve a system of equations with multiple variables, including $c$. However, we need to be careful and make sure that we are using the correct steps and formulas for each equation.

Q: Can we use this method to solve equations with complex numbers or variables?

A: Yes, we can use this method to solve equations with complex numbers or variables. However, we need to be careful and make sure that we are using the correct steps and formulas for each type of number or variable.

Q: What if we have a non-linear equation with a variable $c$? Can we still use this method to solve it?

A: No, we cannot use this method to solve a non-linear equation with a variable $c$. Non-linear equations require a different approach and may involve more complex steps and formulas.

Conclusion

In this article, we answered some frequently asked questions about the equation and its solution. We discussed the significance of the cube root, the importance of assuming that $x > 0$ and $y > 0$, and the possibility of solving the equation for $x$ or $y$ instead of $c$. We also discussed the applicability of this method to different types of equations and variables.