Question

factorise x^6-7x^3-8​

Answer 1

Answer:

Hey There!

${x}^{6} - {7x}^{3} - 8$

Use the rational root theorem

$= ( x- 1) \frac{ {x}^{6} - {7x}^{3} - 8 }{x + 1} \\ \\ = (x + 1)( {x}^{5} - {x}^{4} + {x}^{3} - {8x}^{2} \\ + 8x - 8)$

Factor (x^5-x^4+x^3-8x^2+8x-8)

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

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

$\Large\textsf{Hope \: It \: Helped}$