Question

1. karen wants to prove (sqrt3{8} = 8^{\frac{1}{3}}). which of the following expressions could she use to help her show the two expressions are equivalent? select all that apply. (\bigcirc 2^{\frac{1}{3}} cdot 2^{\frac{1}{3}} cdot 2^{\frac{1}{3}}) (\bigcirc 8^{\frac{1}{3}} cdot 8^{\frac{1}{3}} cdot 8^{\frac{1}{3}}) (\bigcirc (2 cdot 2 cdot 2)^{\frac{1}{3}}) (\bigcirc 8^{\frac{1}{3}} + 8^{\frac{1}{3}} + 8^{\frac{1}{3}}) (\bigcirc (2 + 2 + 2)^{\frac{1}{3}})

Explanation:

To prove \(\sqrt[3]{8}=8^{\frac{1}{3}}\), we can use the property of exponents and radicals. Recall that \(\sqrt[n]{a}=a^{\frac{1}{n}}\) for any real number \(a\) and positive integer \(n\). Also, we know that \(8 = 2^3\).

Step 1: Rewrite \(8\) as a power of \(2\)

We know that \(8 = 2\times2\times2=2^3\). So we can rewrite \(8^{\frac{1}{3}}\) as \((2^3)^{\frac{1}{3}}\)

Step 2: Apply the exponent rule \((a^m)^n=a^{m\times n}\)

For \((2^3)^{\frac{1}{3}}\), using the rule \((a^m)^n = a^{m\times n}\), we have \(m = 3\) and \(n=\frac{1}{3}\), so \((2^3)^{\frac{1}{3}}=2^{3\times\frac{1}{3}}\)

Step 3: Simplify the exponent

\(3\times\frac{1}{3}=1\), so \(2^{3\times\frac{1}{3}} = 2^1=2\)

Now, let's check the value of \(\sqrt[3]{8}\). By definition, \(\sqrt[3]{8}\) is the number \(x\) such that \(x^3 = 8\). We know that \(2^3=8\), so \(\sqrt[3]{8}=2\)

Now let's analyze the given expressions (assuming the expressions are related to showing the equivalence of \(\sqrt[3]{8}\) and \(8^{\frac{1}{3}}\)):

• For an expression like \(8^{\frac{1}{3}}\times8^{\frac{1}{3}}\times8^{\frac{1}{3}}\): Using the rule \(a^m\times a^n=a^{m + n}\), we have \(8^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=8^{\frac{1 + 1+ 1}{3}}=8^{\frac{3}{3}}=8^1 = 8\). But we know that \((\sqrt[3]{8})^3=8\) and \((8^{\frac{1}{3}})^3 = 8\), so this shows that \(8^{\frac{1}{3}}\) cubed is \(8\), which is consistent with the definition of cube root.

• For an expression like \((2\times2\times2)^{\frac{1}{3}}\): Since \(2\times2\times2 = 2^3=8\), then \((2\times2\times2)^{\frac{1}{3}}=(2^3)^{\frac{1}{3}}=2\) (as we calculated before), and \(\sqrt[3]{8}=2\), so this is equivalent.

• For an expression like \(2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}\): \(2^{\frac{1 + 1+ 1}{3}}=2^{\frac{3}{3}}=2^1 = 2\), and \(\sqrt[3]{8}=2\), so this is equivalent.

• For an expression like \((2 - 2)^{\frac{1}{3}}\): \(2-2 = 0\), and \(\sqrt[3]{0}=0

eq2\), so this is not equivalent.

• For an expression like \(8^{\frac{1}{3}}+8^{\frac{1}{3}}+8^{\frac{1}{3}}\): \(8^{\frac{1}{3}}=2\), so \(2 + 2+ 2=6

eq2\), so this is not equivalent.

• For an expression like \(2+2+2+2\): This is \(8

eq2\), so this is not equivalent.

Answer:

The expressions that apply to show \(\sqrt[3]{8}=8^{\frac{1}{3}}\) are the ones that correctly use exponent rules or the definition of cube root to show both equal to \(2\) (or show the consistency of the cube - root and exponent form). For example, if the options include \(8^{\frac{1}{3}}\times8^{\frac{1}{3}}\times8^{\frac{1}{3}}\), \((2\times2\times2)^{\frac{1}{3}}\), \(2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}\) (assuming these are the relevant options), these would be the correct ones. If we assume the options are:

A. \(8^{\frac{1}{3}}\times8^{\frac{1}{3}}\times8^{\frac{1}{3}}\)

B. \((2 - 2)^{\frac{1}{3}}\)

C. \(8^{\frac{1}{3}}+8^{\frac{1}{3}}+8^{\frac{1}{3}}\)

D. \((2\times2\times2)^{\frac{1}{3}}\)

E. \(2+2+2+2\)

F. \(2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}\)

Then the correct options are A, D, F.

(Note: The final answer depends on the exact expressions provided. The above is a general analysis based on the problem of showing \(\sqrt[3]{8}=8^{\frac{1}{3}}\))