a, $x^3 + y^3 + z^3 - 3xyz$
= $(x+y)^3 - 3xy(x+y) + z^3 - 3xyz$
= $[(x+y)^3 + z^3] - [3xy(x+y)+3xyz]$
= $[(x+y)^3 + z^3] - 3xy(x+y+z)$
= $(x+y+z)[(x+y)^2 - (x+y)z + z^2] - 3xy(x+y+z)$
= $(x+y+z)(x^2 + 2xy+y^2-xz-yz+z^2-3xy)$
= $(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$
b, $(x^2+x+1)(x^2+x+2)-12$
= $(x^2+x+1)(x^2+x+1+1)-12$
Đặt $x^2 + x+1 = t$, ta có :
$t(t-1)-12$
= $t^2 + t - 12$
= $t^2 +4t - 3t - 12$
= $(t^2+4t) - (3t + 12)$
= $t(t + 4) - 3(t + 4)$
= $(t+4)(t-3)$
= $(x^2 + x+1 + 4) (x^2+x+1+3)$
= $(x^2+x+5) (x^2 + x+4)$
HỌC TỐT!XIN HAY NHẤT!
