Question

find the following product: (x^2-1) ( x^4+x^2+1)​

Answer 1

Answer:

$\left(x^2-1\right)\left(x^4+x^2+1\right):\quad x^6-1$

Step-by-step explanation:

$\left(x^2-1\right)\left(x^4+x^2+1\right)$

$\gray{\mathrm{Distribute\:parentheses}}$

$=x^2x^4+x^2x^2+x^2\cdot \:1+\left(-1\right)x^4+\left(-1\right)x^2+\left(-1\right)\cdot \:1$

$\gray{\mathrm{Apply\:minus-plus\:rules}}$

$\gray{+\left(-a\right)=-a}$

$=x^4x^2+x^2x^2+1\cdot \:x^2-1\cdot \:x^4-1\cdot \:x^2-1\cdot \:1$

$\gray{\mathrm{Simplify}\:x^4x^2+x^2x^2+1\cdot \:x^2-1\cdot \:x^4-1\cdot \:x^2-1\cdot \:1}$

$x^4x^2+x^2x^2+1\cdot \:x^2-1\cdot \:x^4-1\cdot \:x^2-1\cdot \:1$

$\gray{\mathrm{Group\:like\:terms}}$

$=x^4x^2-1\cdot \:x^4+x^2x^2+1\cdot \:x^2-1\cdot \:x^2-1\cdot \:1$

$\gray{\mathrm{Add\:similar\:elements:}\:1\cdot \:x^2-1\cdot \:x^2=0}$

$=x^4x^2-1\cdot \:x^4+x^2x^2-1\cdot \:1$

$\gray{x^4x^2=x^6}$

$\gray{1\cdot \:x^4=x^4}$

$\gray{x^2x^2=x^4}$

$\gray{1\cdot \:1=1}$

$=x^6-x^4+x^4-1$

$\gray{\mathrm{Add\:similar\:elements:}\:-x^4+x^4=0}$

$=x^6-1$