State which of the following numbers are irrational:

$\begin{matrix} \text{(i)} & 3-\sqrt{\dfrac{7}{25}}\\[1.5em] \text{(ii)} & -\dfrac{2}{3}+\sqrt[3]{2} \\[1.5em] \text{(iii)} & \dfrac{3}{\sqrt{3}} \\[1.5em] \text{(iv)} & -\dfrac{2}{7}\sqrt[3]{5} \\[1.5em] \text{(v)} & (2-\sqrt{3})(2+\sqrt{3}) \\[1.5em] \text{(vi)} & (3+\sqrt{5})^2 \\[1.5em] \text{(vii)} &(\dfrac{2}{5}\sqrt{7})^2 \\[1.5em] \text{(viii)} & (3-\sqrt{6})^2 \\[1.5em] \end{matrix}$

$\text{(i) } 3 - \sqrt{\dfrac{7}{25}} = 3 - \dfrac{\sqrt7}{(\sqrt{5 × 5})} = 3 - \dfrac{\sqrt7}{5}$

As , $3 - \dfrac{\sqrt{7}}{5}$ is an irrational number,

∴ $\bold{3 - \sqrt{\dfrac{7}{25}}}$ is also an irrational number.

$\text{(ii) } -\dfrac{2}{3} + \sqrt[3]{2}$

Here, 2 is not perfect cube

∴ $-\dfrac{2}{3} + \sqrt[3]{2}$ is an irrational number.

$\text{(iii) } \dfrac{3}{\sqrt{3}} = \dfrac{3}{\sqrt{3}} × \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{3\sqrt{3}}{3} = \sqrt{3}$

As, $\sqrt{3}$ is an irrational number.

∴ $\dfrac{3}{\sqrt{3}}$ is an irrational number.

$\text{(iv) } -\dfrac{2}{7}\sqrt[3]{5}$

Here, 5 is not perfect cube

∴ $-\dfrac{2}{7}\sqrt[3]{5}$ is an irrational number.

$\text{(v) } (2-\sqrt{3})(2+\sqrt{3})$

Using identity : $(a + b)(a - b) = a^2 - b^2$

$(2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt3)^2 = 4 - 3 = 1$

Hence , $(2-\sqrt{3})(2+\sqrt{3})$ is a rational number.

$\text{(vi) }(3 + \sqrt{5})^2$

Using identity : (a + b)² = a² + 2ab + b²

$(3 + \sqrt{5})^2 = 3^2 + 2 × 3 × \sqrt{5} + (\sqrt{5})^2 \\[1.5em] = 9 + 6\sqrt{5} + 5 \\[1.5em] = 9 + 5 + 6\sqrt{5} \\[1.5em] = 14 + 6\sqrt{5}$

As, 14 + $6\sqrt{5}$ is an irrational number

∴ $(3+\sqrt{5})^2$ is an irrational number.

$\text{(vii) } \Big(\dfrac{2}{5}\sqrt{7}\Big)^2$

$\Big(\dfrac{2}{5}\sqrt{7}\Big)^2 = \dfrac{2}{5}\sqrt{7} × \dfrac{2}{5}\sqrt{7} \\[1.5em] = \dfrac{4}{25} × (\sqrt7)^2 \\[1.5em] = \dfrac{4}{25} × 7 \\[1.5em] = \dfrac{28}{25}$

As, $\dfrac{28}{25}$ is a rational number,

∴ $(\dfrac{2}{5}\sqrt{7})^2$ is a rational number.

$\text{(viii) } (3 - \sqrt{6})^2$

Using identity : (a + b)² = a² + 2ab + b²

$(3 - \sqrt{6})^2 = 3^2 - 2 × 3 × \sqrt{6} + (\sqrt{6})^2 \\[1.5em] = 9 - 6\sqrt{6} + 6 \\[1.5em] = 9 + 6 - 6\sqrt{6} \\[1.5em] = 15 - 6\sqrt{6}$

As, 15 - $6\sqrt{6}$ is an irrational number.

∴ $(3 - \sqrt{6})^2$ is an irrational number.