Problem 37
Question
Simplify \(\left(x^{4} y^{0} z^{2}\right)^{3}\)
Step-by-Step Solution
Verified Answer
Question: Simplify the expression \((x^4y^0z^2)^3\).
Answer: \(x^{12}z^6\)
1Step 1: Apply the power rule
The power rule states that \((a^m)^n = a^{mn}\). Apply this rule to each term inside the parentheses: \((x^4)^3\), \((y^0)^3\), and \((z^2)^3\).
2Step 2: Calculate the powers
Using the power rule from the first step, we get:
\(x^{(4\cdot3)}\),
\(y^{(0\cdot3)}\),
\(z^{(2\cdot3)}\).
Now, calculate the powers:
\(x^{12}\),
\(y^0\),
\(z^6\).
3Step 3: Simplify the expression
Now, simplify the expression by multiplying the calculated terms. Keep in mind that any number raised to the power of 0 is equal to 1. The simplified expression is:
\(x^{12} \cdot y^0 \cdot z^6\).
Since \(y^0 = 1\), our final simplified expression is:
\(x^{12} \cdot 1 \cdot z^6 \Rightarrow x^{12}z^6\).
The simplified expression is \(\boxed{x^{12}z^6}\).
Key Concepts
ExponentiationSimplificationPower Rule
Exponentiation
Exponentiation refers to the mathematical operation involving two numbers, the base and the exponent. The exponent tells us how many times to multiply the base by itself. In the expression \(x^4\), \(x\) is the base, and \(4\) is the exponent. This means you multiply \(x\) by itself four times: \(x \cdot x \cdot x \cdot x\).
Exponentiation is a fundamental concept in algebra because it helps to express repeated multiplication compactly. When an expression like \((x^4 y^0 z^2)^3\) is encountered, the exponentiation applies not only to each base individually, but also to the entire expression inside the parentheses. Therefore, each term needs to be evaluated with the exponent outside the parentheses.
Exponentiation is a fundamental concept in algebra because it helps to express repeated multiplication compactly. When an expression like \((x^4 y^0 z^2)^3\) is encountered, the exponentiation applies not only to each base individually, but also to the entire expression inside the parentheses. Therefore, each term needs to be evaluated with the exponent outside the parentheses.
Simplification
Simplification in mathematics is the process of reducing a complex expression into a simpler, more manageable form. This often means combining like terms, reducing fractions, or resolving powers of zero. In our case, simplifying the expression \((x^4 y^0 z^2)^3\) involves applying the rules of exponents and reducing it to its simplest form.
One important thing to remember during simplification is understanding and applying the rule \(a^0 = 1\). It explains why \(y^0 = 1\), allowing us to simplify the expression by removing \(y^0\) since multiplying any term by 1 leaves it unchanged. Thus, the expression \((x^4 y^0 z^2)^3\) gets reduced to \(x^{12}z^6\) after applying the rules correctly.
Simplification makes calculations easier, especially when dealing with complex expressions, by breaking them down into manageable pieces.
One important thing to remember during simplification is understanding and applying the rule \(a^0 = 1\). It explains why \(y^0 = 1\), allowing us to simplify the expression by removing \(y^0\) since multiplying any term by 1 leaves it unchanged. Thus, the expression \((x^4 y^0 z^2)^3\) gets reduced to \(x^{12}z^6\) after applying the rules correctly.
Simplification makes calculations easier, especially when dealing with complex expressions, by breaking them down into manageable pieces.
Power Rule
The power rule is a key property of exponents that allows us to simplify expressions involving powers of powers. It states that \((a^m)^n = a^{m \cdot n}\). This means when you have an exponent raised to another exponent, you multiply the exponents.
In the exercise, the power rule was applied to simplify each piece of \((x^4 y^0 z^2)^3\). Here is how it works step-by-step:
In the exercise, the power rule was applied to simplify each piece of \((x^4 y^0 z^2)^3\). Here is how it works step-by-step:
- For \(x^4\) raised to the third power, apply the rule: \((x^4)^3 = x^{4 \times 3} = x^{12}\).
- For \(y^0\) raised to the third power: \((y^0)^3 = y^{0 \times 3} = y^0\).
- For \(z^2\) raised to the third power: \((z^2)^3 = z^{2 \times 3} = z^6\).
Other exercises in this chapter
Problem 37
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