simplify: index of 3 √y(2y index of √8y^2- index of 3 √y^5-4 index of 3√8 y^2

1 answer

To simplify the expression you provided:

\[ \text{index of } 3 \sqrt{y} \left(2y \text{ index of } \sqrt{8y^2} - \text{ index of } 3 \sqrt{y^5} - 4 \text{ index of } 3 \sqrt{8y^2}\right) \]

We need to break this down step by step.

  1. Understanding the expression:

    • The expression is not completely clear because it contains unspecified indices and my notation could be misinterpreted.
    • I will interpret them as follows: regarding the "index" of the root, I'll treat "index of \(n \sqrt{x}\)" as referring to the \(n\)-th root, which is \(x^{1/n}\).
  2. Rewrite the roots as powers:

    • \(3 \sqrt{y} = y^{1/3}\)
    • \(\sqrt{8y^2} = (8y^2)^{1/2} = 4y\)
    • Usual representation of \(3 \sqrt{y^5}\) is \(y^{5/3}\)
    • \(3 \sqrt{8y^2} = (8y^2)^{1/3} = 2^{5/3}y^{2/3}\)

Now let's rewrite the entire expression:

\[ y^{1/3} \left(2y \cdot 4y - y^{5/3} - 4 \cdot 2^{5/3} y^{2/3}\right) \]

  1. Calculate each component:

    • \(2y \cdot 4y = 8y^2\)
    • Thus, the expression becomes:

\[ y^{1/3} \left(8y^2 - y^{5/3} - 4 \cdot 2^{5/3} y^{2/3}\right) \]

  1. Distributing \(y^{1/3}\):

    • Distributing \(y^{1/3}\) through: \[ y^{1/3} \cdot 8y^2 - y^{1/3} \cdot y^{5/3} - y^{1/3} \cdot 4 \cdot 2^{5/3} y^{2/3} \]

    • This becomes:

      • \(8y^{2 + 1/3} = 8y^{7/3}\)
      • \(- y^{(1/3 + 5/3)} = - y^{6/3} = -y^2\)
      • \(-4 \cdot 2^{5/3} y^{(1/3 + 2/3)} = -4 \cdot 2^{5/3} y^{1} = -4 \cdot 2^{5/3} y\)
  2. Combine everything:

Putting it all together we have:

\[ 8y^{7/3} - y^2 - 4 \cdot 2^{5/3} y \]

  1. Final expression:

Thus the simplified expression is:

\[ 8y^{7/3} - y^2 - 4 \cdot 2^{5/3} y \]

This is a fully simplified form given the initial interpretation based on your query. If anything needs adjustment or does not align with your intended meaning, please clarify!