Question

Simplify the following products:

$(x^3-3x^2-x)(x^2-3x+1)$

Answer 1

Given,$(x^3-3x^2-x)(x^2-3x+1)$
$=x(x^2-3x-1)(x^2-3x+1)$
$=x\left(\right(x^2-3x)-1)\left(\right(x^2-3x)+1)$
using identity $(a-b)(a+b)=a^2-b^2$
$=x\left(\right(x^2-3x{)}^2-1^2)$
using identity, $(a-b{)}^2=a^2-2ab+b^2$
$=x\left(\right(x^2{)}^2-2\times x^2\times 3x+(3x{)}^2-1^2)$
$=x(x^4-6x^3+9x^2-1)$

Hence, $(x^3-3x^2-x)(x^2-3x+1)=x(x^4-6x^3+9x^2-1).$