Answer:
[tex] \displaystyle{ \frac{x + 4}{x - 4} + \frac{x - 4}{x + 4} = 3 \frac{1}{3} }[/tex]
[tex]\displaystyle{ \frac{ {(x + 4) }^{2} + {(x - 4)}^{2} }{(x + 4)(x - 4)} = \frac{10}{3} }[/tex]
[tex]\displaystyle{ \frac{ {x}^{2} + 8x + {4}^{2} + {x}^{2} - 8x + {4}^{2} }{ {x}^{2} - {4}^{2} } = \frac{10}{3} }[/tex]
[tex]\displaystyle{ \frac{ {2x}^{2} + 16 + 16 }{ {x}^{2} - 16 } = \frac{10}{3} }[/tex]
[tex]\displaystyle{3( {2x}^{2} + 32) = 10( {x}^{2} - 16) }[/tex]
[tex]\displaystyle{ {6x}^{2} + 96 = {10x}^{2} - 160 }[/tex]
[tex]\displaystyle{ {10x}^{2} - 160 = {6x}^{2} + 96 }[/tex]
[tex]\displaystyle{ {10x}^{2} - {6x}^{2} - 160 - 96 = 0 }[/tex]
[tex]\displaystyle{ {4x}^{2} - 256 = 0}[/tex]
[tex]\displaystyle{4( {x}^{2} - 64) = 0}[/tex]
[tex] {x}^{2} - {8}^{2} = \displaystyle{ \frac{0}{4} }[/tex]
[tex]\displaystyle{(x + 8)(x - 8) = 0}[/tex]
[tex]\displaystyle{ \rm{either, \: x + 8 = 0............(1) }}[/tex]
[tex] \rm{or ,\: x - 8 = 0...............(2)}[/tex]
From equation (1)
[tex]x + 8 = 0[/tex]
[tex]x = - 8[/tex]
From equation (2)
[tex]x - 8 = 0[/tex]
[tex]\displaystyle{x = 8}[/tex]
[tex] \therefore{x=±8}[/tex]
