Question

x³ + 8
27 + a³
1 + m^6
p^9 + 64

Answer 1

$displaystyle tt x^3+8=x^3+2^3=(x+2)(x^2-2x+2^2)=boxed{bold{(x+2)(x^2-2x+4)}}$

$displaystyle tt 27+a^3=3^3+a^3=(3+a)(3^2-3a+a^2)=boxed{bold{(3+a)(9-3a+a^2)}}$

$displaystyle tt 1+m^6=(1+m^2)(1-1m^2+m^4)=boxed{bold{(1+m^2)(1-m^2+m^4)}}$

$displaystyle tt p^9+64=p^{3cdot3}+4^3=(p^3)^3+4^3=(p^3+4)((p^3)^2-p^3cdot4+4^2=boxed{bold{(p^3+4)(p^6-4p^3+16)}}$

Answer 2

Ответ:

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

$2.(3 + a)(9 - 3a + {a}^{2} )$

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

$4.(p {}^{3} + 4)(p {}^{6} - 4 {p}^{3} + 16)$