Question

If (x²-1)\x=4 then find (x^6-1)/x³

Answer 1

$\frac{ {x}^{2} - 1 }{x} = 4 \\ taking \: cube \: on \: both \: side \\ {( \frac{ {x}^{2} - 1 }{x}) }^{3} = {4}^{3} \\ \frac{ {x}^{6 } - 1 - 3 {x}^{4 } + 3 {x}^{2} }{ {x}^{3} } = 64 \\ \frac{ {x}^{6} - 1 - 3 {x}^{2}( {x}^{2} - 1) } { {x}^{3} } = 64\\ \frac{ {x}^{6} - 1 }{ {x}^{3} } - \frac{3 {x}^{2}( {x}^{2} - 1) }{ {x}^{3} } = 64 \\ \frac{ {x}^{6} - 1}{ {x}^{3} } - 12 = 64 \\ \frac{ {x}^{6} - 1}{ {x}^{3} } = 64 + 12 \\ \frac{ {x}^{6} - 1 }{ {x}^{3} } = 76$

The step by step explanation