Question

How do I factorise x6+4x3-1

Answer 1

Hey

Let's just start Factorising :

${x}^{6} + 4 {x}^{3} - 1$

$= {( {x}^{2} )}^{3} + {x}^{3} - 1 - 3( {x}^{2} )(x)( - 1)$

By formula for Factorising :

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)$

Use this to get the factor of above as :

$( {x}^{2} + x - 1)( {x}^{4} + {x}^{2} + 1 - {x}^{3} + x + {x}^{2} )$

Arranging :

$( {x}^{2} + x - 1)( {x}^{4} - {x}^{3}+ 2{x}^{2} + x + 1 )$

Well, no factors further -_-