Question

if X square + 1 upon 4 x square equals to 8 then find 8 x cube + 1 upon 8 x cube​

Answer 1

${x}^{2} + \frac{1}{4 {x}^{2} } = 8 \\ {x}^{2} + 4 {x}^{ - 2} = 8 \\ {x }^{2} + {x}^{ - 2} = \frac{8}{4} \\ {x}^{2} + {x}^{ - 2} = 2 \\ x + {x}^{ - 2} = \sqrt{2} \\ x + \frac{1}{ {x}^{2} } = \sqrt{2} \\ \frac{ {x}^{3} + 1}{ {x}^{2} } = \sqrt{2} \\ \frac{ {x}^{3} }{ {x}^{2} } = \frac{ \sqrt{2} }{1} \\ x = \sqrt{2}$
find value you are smart.

Answer 2

Answer:

Step-by-step explanation: