State, with reason, which of the following are surds and which are not :

(i) $\sqrt{180}$

(ii) $\sqrt[4]{27}$

(iii) $\sqrt[5]{128}$

(iv) $\sqrt[3]{64}$

(v) $\sqrt[3]{25}.\sqrt[3]{40}$

(vi) $\sqrt[3]{-125}$

(vii) $\sqrt{π}$

(viii) $\sqrt{3 + \sqrt{2}}$

(i) Given,

$\sqrt{180} = \sqrt{2 \times 2 \times 3 \times 3 \times 5} = 2 \times 3 \times \sqrt{5} = 6\sqrt{5}$, which is an irrational number.

Since, 180 is a rational number and $\sqrt{180}$ is an irrational number.

Hence, $\sqrt{180}$ is a surd.

(ii) Given,

$\sqrt[4]{27} = \sqrt[4]{3^3} = (3)^{\dfrac{3}{4}}$, which is irrational.

Since, 27 is a rational number and $\sqrt[4]{27}$ is an irrational number.

Hence, $\sqrt[4]{27}$ is a surd.

(iii) Given,

$\sqrt[5]{128} = \sqrt[5]{2^7} = \sqrt[5]{2^5 \times 2^2} \\[1em] = (2^5)^{\frac{1}{5}} \times (2^2)^{\frac{1}{5}} \\[1em] = 2 \times (2)^{\frac{2}{5}}$

which is irrational.

Since, 128 is a rational number and $\sqrt[5]{128}$ is an irrational number.

Hence, $\sqrt[5]{128}$ is a surd.

(iv) Given,

$\sqrt[3]{64} = \sqrt[3]{4^3} = 4^{3 \times \dfrac{1}{3}}$ = 4, which is rational.

Since, 64 is rational and $\sqrt[3]{64}$ is also rational.

Hence, $\sqrt[3]{64}$ is not a surd.

(v) Given,

$\sqrt[3]{25}.\sqrt[3]{40} = \sqrt[3]{1000} = \sqrt[3]{10^3} = 10^{3 \times \dfrac{1}{3}}$ = 10, which is rational.

Since, 1000 is rational and 10 is also rational.

Hence, $\sqrt[3]{25}.\sqrt[3]{40}$ is not a surd.

(vi) Given,

$\sqrt[3]{-125} = \sqrt[3]{(-5)^3} = (-5)^{3 \times \dfrac{1}{3}} = -5$, which is rational,

Since, -125 is rational and -5 is also rational.

Hence, $\sqrt[3]{-125}$ is not a surd.

(vii) Given,

$\sqrt{π}$

Since, π is irrational and $\sqrt{π}$ is also irrational.

Hence, $\sqrt{π}$ is not a surd.

(viii) Given,

$\sqrt{3 + \sqrt{2}}$

Since, $3 + \sqrt{2}$ is irrational.

Hence, $\sqrt{3 + \sqrt{2}}$ is not a surd.