[tex]\text{Given that,}\\\\x-\dfrac 3x = 7\\\\\\\textbf{i).}\\\\\\x^2 + \dfrac 9{x^2}\\\\\\=x^2 + \left(\dfrac 3 x \right)^2\\\\\\=\left(x- \dfrac 3x \right)^2 + 2 \cdot x \cdot \dfrac 3x\\\\\\=7^2 +6\\\\\\=55[/tex]
[tex]\textbf{ii)}\\\\~~~~~\left(x+\dfrac 3 x \right)^2 = \left( x - \dfrac 3x \right)^2+4 \cdot x \cdot \dfrac 3x\\\\\\\implies \left(x+\dfrac 3 x \right)^2 = 7^2 +12\\\\\\\implies \left(x+\dfrac 3 x \right)^2 =61\\\\\\\implies x+ \dfrac 3x = \pm\sqrt{61}\\\\\\[/tex]
[tex]\textbf{iii)}\\\\\\x^2 -\dfrac 9{x^2}\\\\\\=x^2 - \left(\dfrac 3{x} \right)^2\\\\\\=\left( x-\dfrac 3x \right)\left( x+\dfrac 3x \right)\\\\\\=7\sqrt{61}~~ \text{or}~~ -7\sqrt{61}[/tex]
