Question

DeltaMath - Combine Radicals/Fractional Exponents (L2)

I don't understand how to solve this type of problem. Help needed

Answer 1

`\frac{ \sqrt[3]{27x {y}^(3) } }{5 {x}^{ (4)/(3) } {y}^(2) } = \frac{3 {x}^{ (1)/(3) }y }{5 {x}^{ (4)/(3) } {y}^(2) } = (3)/(5) {x}^( - 1) {y}^( - 1)`

`abc = (3)/(5) ( - 1)( - 1) = (3)/(5)`

Answer 2

The expression simplifies to
`\((3)/(5xy)\)`. In the form
`\(ax^by^c\)`, the product of a, b, and c is
`\((3)/(5)\)`.

Let's simplify the given expression:

`\[ \frac{\sqrt[3]{27xy^3}}{5x^(4/3)y^2} \]`

First, simplify the numerator:

`\[ \sqrt[3]{27xy^3} = \sqrt[3]{3^3 \cdot (xy^3)} = 3\sqrt[3]{xy^3} \]`

Now substitute this back into the original expression:

`\[ \frac{3\sqrt[3]{xy^3}}{5x^(4/3)y^2} \]`

Now, let's simplify the expression further. Combine the terms in the numerator and simplify the exponents:

`\[ (3x^(1/3)y)/(5x^(4/3)y^2) \]`

Combine the x terms by subtracting the exponents:

`\[ (3)/(5) \cdot (1)/(x^(4/3-1/3)) \cdot (y)/(y^2) \]`

Simplify the exponents:

`\[ (3)/(5) \cdot (1)/(x) \cdot (1)/(y) \]`

Now, write this expression in the form
`\( ax^by^c \)`:

`\[ (3)/(5xy) \]`

Now, compare this with the desired form
`\( ax^by^c \)`:

`\[ a = (3)/(5), \quad b = -1, \quad c = -1 \]`

The product of a, b, and c is:

`\[ abc = (3)/(5) \cdot (-1) \cdot (-1) = (3)/(5) \]`