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What value of \( c \) makes the equation true? Assume \( x>0 \) and \( y>0 \) \( \sqrt[3]{\frac{x^{3}}{c y^{4}}}=\frac{x}{4 y(\sqrt[3]{y})} \) \( c=12 \) \( c=16 \) \( c=64 \) \( c=81 \)

Ask by Chang Higgins. in the United States
Feb 25,2025

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\( c = 64 \)

Solución

We start with the equation   ∛(x³/(c·y⁴)) = x/(4y·∛y). Since x > 0 and y > 0, we can safely cube both sides:   (x³/(c·y⁴)) = [x/(4y·∛y)]³. Expanding the right side, note that   (x/(4y·∛y))³ = x³ / (4³·y³·(∛y)³). Since (∛y)³ = y, it follows that   (x/(4y·∛y))³ = x³ / (64·y³·y) = x³ / (64·y⁴). Now setting the cube of the left side equal to the cube of the right side:   x³/(c·y⁴) = x³/(64·y⁴). Because x³ and y⁴ are positive, we can cancel them from both sides:   1/c = 1/64  ⇒ c = 64. Thus, the value of c that makes the equation true is 64.

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To find the appropriate value of \( c \) that makes the equation true, we can start by simplifying both sides of the equation. Starting with the left side: \[ \sqrt[3]{\frac{x^{3}}{c y^{4}}} = \frac{x}{\sqrt[3]{c} \cdot \sqrt[3]{y^4}} = \frac{x}{\sqrt[3]{c} \cdot y^{\frac{4}{3}}} \] Now, simplifying the right side: \[ \frac{x}{4 y(\sqrt[3]{y})} = \frac{x}{4y^{\frac{4}{3}}} \] Setting both sides equal gives: \[ \frac{x}{\sqrt[3]{c} \cdot y^{\frac{4}{3}}} = \frac{x}{4y^{\frac{4}{3}}} \] Dividing both sides by \( x \) (assuming \( x > 0 \)), we have: \[ \frac{1}{\sqrt[3]{c}} = \frac{1}{4} \] By cross-multiplying, we find: \[ \sqrt[3]{c} = 4 \] Now, cubing both sides leads us to: \[ c = 4^3 = 64 \] Thus, the value of \( c \) that makes the equation true is \( 64 \). Now, for some fun facts: Did you know that cubes and their roots are a huge part of geometry? The volume of a cube is calculated by raising the length of one side to the third power! That's why you see cubed terms popping up in formulas everywhere, especially in calculus! Also, in real-life scenarios, understanding these principles can help entrepreneurs. If you're working with materials and need to optimize space in your storage area, knowing how to manipulate volume calculations can lead you to success. So whether it’s maximizing your warehouse or planning a new product, having fun with cubes can yield great results!

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