Problem 40 - Intermediate Algebra

Question

Multiply and simplify. All variables represent positive real numbers. $$ (4 \sqrt[3]{9}-3 \sqrt[3]{3})(4 \sqrt[3]{3}+2 \sqrt[3]{6}) $$

Step-by-Step Solution

Answer

The simplified expression is \(3^{2/3}(4\sqrt[3]{3} - 3)(4\sqrt[3]{3} + 3)\).

Step 1: Recognize the Binomial Form

Observe that the given expression \((4 \sqrt[3]{9}-3 \sqrt[3]{3})(4 \sqrt[3]{3}+2 \sqrt[3]{6})\) is in the form of \((a-b)(a+b)\). In such cases, we use the formula for the difference of squares: \((a-b)(a+b) = a^2 - b^2\).

Step 2: Identify a and b

Identify \(a\) and \(b\) from the expression.Here, \(a = 4\sqrt[3]{9}\) and \(b = 3\sqrt[3]{3}\).

Step 3: Calculate a²

Calculate the square of \(a\):\[a^2 = (4\sqrt[3]{9})^2 = 16 \times (9^{2/3}) = 16 \times (9^{2/3}) = 16 \times (3^2)^\frac{2}{3}= 16 \times 3^{4/3} = 16 \times 3^{4/3}\]

Step 4: Calculate b²

Calculate the square of \(b\):\[b^2 = (3\sqrt[3]{3})^2 = 9 \times (3^{2/3}) = 9 \times 3^{2/3}= 9 \times 3^{2/3} = 9 \times 3^{2/3}\]

Step 5: Apply the Difference of Squares

Apply the difference of squares formula:\[a^2 - b^2 = 16 \times 3^{4/3} - 9 \times 3^{2/3}\]

Step 6: Simplify the Expression

To simplify the expression, let's rewrite it using a common base:Let \(x = 3^{1/3}\), then:\(a^2 = 16x^4\) and \(b^2 = 9x^2\).Now, the expression becomes:\[16x^4 - 9x^2\]Factor out the greatest common factor, \(x^2\):\[x^2(16x^2 - 9)\]Realize that \(16x^2 - 9\) is a difference of squares:\[x^2((4x)^2 - 3^2) = x^2(4x - 3)(4x + 3)\]Substituting back, \(x = 3^{1/3}\):\[3^{2/3}(4\cdot3^{1/3} - 3)(4\cdot3^{1/3} + 3)\]

Step 7: Final Expression

Further simplifying gives us the expression as:\[3^{2/3}(4\cdot3^{1/3} - 3)(4\cdot3^{1/3} + 3)\]Simplifying the terms:\[= 3^{2/3}(4\sqrt[3]{3} - 3)(4\sqrt[3]{3} + 3)\]

Key Concepts

Binomial Expressions

A binomial expression is made up of two terms separated by a plus or minus sign. Think of it as a small package containing a pair of different items, for example,

• \(x + y\)
• \(a - b\)

In the given exercise, our binomial expressions are

• \(4 \sqrt[3]{9} - 3 \sqrt[3]{3}\)
• \(4 \sqrt[3]{3} + 2 \sqrt[3]{6}\)

Each of these binomials can consist of constants, variables, or a combination of both. Finding a common pattern in binomials can help simplify complex expressions, especially when multiplying them.

Difference of Squares

The difference of squares is a special pattern with the formula:\[(a-b)(a+b) = a^2 - b^2\]This is like unraveling a magic trick where the middle terms cancel each other.

In our exercise, the expression

• \((4\sqrt[3]{9}-3\sqrt[3]{3})(4\sqrt[3]{3}+2\sqrt[3]{6})\)

looks quite complex, but once it's recognized to fit the difference of squares, simplifying becomes much easier.

By identifying the terms correctly as \(a\) and \(b\), we can use this powerful technique to reduce the problem to something manageable.

Radicals and Exponents

Radicals and exponents are like siblings in mathematics, both dealing with powers and roots, but from opposite sides. A radical, such as a cube root \(\sqrt[3]{x}\), can be rewritten using exponents: \(x^{1/3}\).

In our exercise, there's a blend of radicals and exponents. Understanding them:

• The term \( \sqrt[3]{9} \) can be expressed as \( 9^{1/3} \).
• Multiplying radicals involves adding their exponents when they're in base form.

By conducting calculations in their base and exponent form, we can more easily perform operations such as squaring the terms, which flows into the simplification process.

Simplification of Algebraic Expressions

Simplification is all about making expressions smaller and tidier while preserving their value. Think of it as organizing cluttered rooms to be neat and efficient. Our goal is to spot patterns and forms that can help reduce complexity.

In our solution, simplification involved several steps:

• Using the difference of squares pattern to eliminate extra terms.
• Rewriting radicals in terms of exponents to enable easier multiplication and division.
• Factoring out common terms to make the expression as simple as possible.

By applying these strategies systematically, one can solve even intricate algebraic problems efficiently and with confidence.