Problem 37
Question
\(29-38=\) Simplify the expression. Assume that the letters denote any real numbers. $$ \sqrt[3]{\sqrt{64 x^{6}}} $$
Step-by-Step Solution
Verified Answer
The simplified expression is \(2x\).
1Step 1: Evaluate the Inner Expression
First, observe the expression under the cube root: \(\sqrt{64 x^6}\). Simplify the inner square root: \(\sqrt{64}\) simplifies to \(8\), and \(\sqrt{x^6}\) simplifies to \(x^3\) because whenever you take the square root of an expression raised to a power, you divide the exponent by two. Therefore, \(\sqrt{64 x^{6}} = 8x^3\).
2Step 2: Substitute Back into the Cube Root
Next, substitute the simplified expression back: \(\sqrt[3]{8x^3}\).
3Step 3: Evaluate the Cube Root
Now, simplify the cube root. The expression \(8x^3\) inside the cube root can be rewritten as \((2x)^3\) since \(8 = 2^3\) and \(x^3\) can be directly observed. Thus, we have \(\sqrt[3]{(2x)^3}\). The cube root and the cube cancel each other, leading to the result \(2x\).
4Step 4: Conclusion
The simplified expression is \(2x\). Therefore, the cube root of the square root of \(64 x^6\) simplifies to \(2x\).
Key Concepts
Cube RootSquare RootExponentsAlgebraic Simplification
Cube Root
The cube root is a mathematical function that reverses the operation of cubing a number. When we talk about the cube root of a number, we're looking for a value that, when multiplied by itself three times, gives us the original number. For example, the cube root of 27 is 3, because \(3 \times 3 \times 3 = 27\).
In the exercise, we had the expression \(\sqrt[3]{8x^3}\). This can be rewritten as \((2x)^3\), because 8 is equal to \(2^3\), which allows the cube and cube root to effectively "cancel" each other out, resulting in the simplified expression \(2x\).
In the exercise, we had the expression \(\sqrt[3]{8x^3}\). This can be rewritten as \((2x)^3\), because 8 is equal to \(2^3\), which allows the cube and cube root to effectively "cancel" each other out, resulting in the simplified expression \(2x\).
- Cube roots can be calculated for both positive and negative numbers. For example, \(\sqrt[3]{-8} = -2\).
- Cube roots are particularly useful in algebraic simplification where expressions involve powers of three.
Square Root
The square root is a fundamental concept in algebra. It involves finding a number which, when multiplied by itself, returns the original number. An example is the square root of 64, which is 8 because \(8 \times 8 = 64\).
In the given exercise, we began with the square root \(\sqrt{64x^6}\). This separated into two components: \(\sqrt{64}\) and \(\sqrt{x^6}\).
In the given exercise, we began with the square root \(\sqrt{64x^6}\). This separated into two components: \(\sqrt{64}\) and \(\sqrt{x^6}\).
- \(\sqrt{64}\) simplifies to 8.
- \(\sqrt{x^6}\) simplifies to \(x^3\), achieved by dividing the exponent by two.
Exponents
In mathematics, exponents are a way to represent repeated multiplication. For instance, \(x^3\) means \(x \times x \times x\). Exponents have specific rules that simplify the manipulation of expressions, such as when simplifying square and cube roots.
A critical rule when dealing with square roots is that the power or exponent of the number under the root is divided by two. For example, with \(x^6\), the square root becomes \(x^3\). Similarly, when taking cube roots, you divide the exponent by three.
A critical rule when dealing with square roots is that the power or exponent of the number under the root is divided by two. For example, with \(x^6\), the square root becomes \(x^3\). Similarly, when taking cube roots, you divide the exponent by three.
- Simplifying \(8x^3\) involves breaking it into base components: Whoa, \((2^3) \times (x^3)\), which allows direct extraction under a cube root.
- This process ensures accurate simplification, leading to the correct final result of \(2x\).
Algebraic Simplification
Algebraic simplification is about making an expression as simple as possible. It involves using fundamental rules and properties to combine like terms and reduce expressions to their minimal forms.
In the problem we encountered, algebraic simplification began with breaking down compounds like \(\sqrt{64x^6}\). By applying rules of square roots and exponents, we could transform the expression into a more manageable form, \(8x^3\), followed by further simplification under a cube root.
In the problem we encountered, algebraic simplification began with breaking down compounds like \(\sqrt{64x^6}\). By applying rules of square roots and exponents, we could transform the expression into a more manageable form, \(8x^3\), followed by further simplification under a cube root.
- Each step aims to reduce complexity, helping not only in solving but also understanding the algebraic functions behind expressions.
- Being efficient in simplification requires practice with the manipulation of algebraic rules, especially with roots and powers.
Other exercises in this chapter
Problem 37
\(35-82\) Factor the expression completely. $$ 6 y^{4}-15 y^{3} $$
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\(33-38\) . Express the interval in terms of inequalities, and then graph the interval. $$ [2, \infty) $$
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Multiply the algebraic expressions using the FOIL method, and simplify. \((x+4)(x-3)\)
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Perform the indicated operations. \(\frac{2}{\frac{2}{3}}-\frac{\frac{2}{3}}{2}\)
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