Question

determine whether the numbers in the first column are rational or irrational.
select rational numbers or irrational numbers.
row 1
$(sqrt{2})^8$
row 2
$sqrt3{6}$
row 3
$sqrt3{27}$
row 4
$3sqrt{18}$

Explanation:

Step1: Simplify $(\sqrt[3]{2})^6$

Use exponent rule: $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$. So $(\sqrt[3]{2})^6 = 2^{\frac{6}{3}} = 2^2 = 4$

Step2: Classify $(\sqrt[3]{2})^6$

4 is integer, so rational.

Step3: Analyze $\sqrt[3]{6}$

6 has no perfect cube factors, so $\sqrt[3]{6}$ cannot be simplified to integer/fraction.

Step4: Classify $\sqrt[3]{6}$

Non-integer/fraction, so irrational.

Step5: Simplify $\sqrt[3]{27}$

$\sqrt[3]{27} = \sqrt[3]{3^3} = 3$

Step6: Classify $\sqrt[3]{27}$

3 is integer, so rational.

Step7: Analyze $3\sqrt[3]{18}$

18 has no perfect cube factors, so $\sqrt[3]{18}$ is irrational; multiplying by 3 keeps it irrational.

Step8: Classify $3\sqrt[3]{18}$

Non-integer/fraction, so irrational.

Answer:

Row 1: $(\sqrt[3]{2})^6$: Rational Number
Row 2: $\sqrt[3]{6}$: Irrational Number
Row 3: $\sqrt[3]{27}$: Rational Number
Row 4: $3\sqrt[3]{18}$: Irrational Number