Answer
$(x+1)^{3}=x^{3}+3x^{2}+3x+1$
$(x+2)^{3}=x^{3}+6x^{2}+12x+8$
$(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$
Work Step by Step
$(x+1)^{3}=(x+1)^{2}(x+1)$
$=(x^{2}+2x+1)(x+1)$
$=x^{3}+x^{2}+2x^{2}+2x+x+1$
$=x^{3}+3x^{2}+3x+1$
$(x+2)^{3}=(x+2)^{2}(x+2)$
$=[x^{2}+2(x)(2)+2^{2}](x+2)$
$=(x^{2}+4x+4)(x+2)$
$=x^{3}+2x^{2}+4x^{2}+8x+4x+8$
$=x^{3}+6x^{2}+12x+8$
We infer from the above two expansions that
$(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$