Respuesta :

apologiabiology
group and factor
trial and error

ok, so we group
(x⁵+x³)+(x²+1)
factor out gcf from each
x³(x²+1)+(x²+1)
undistribute
(x²+1)(x³+1)
factor sum of 2  perfect cubes
remember that a³+b³=(a+b)(a²-ab+b²)
so
(x²+1)(x+1)(x²-x+1) is factored form