Algebra and Applications Test 2

Concept:Use the algebraic identity: if $a+b+c=0$, then $a^3+b^3+c^3=3abc$.Explanation:Let $a=x-1$, $b=y-2$, and $c=z-3$.Then $a+b+c=(x-1)+(y-2)+(z-3)$.So $a+b+c=x+y+z-6$.Given $x+y+z=6$, we get $a+b+c=6-6=0$.Since the sum of $a$, $b$, and $c$ is zero, the cubic identity applies directly.This identity follows from the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.When $a+b+c=0$, the right side becomes $0$, so $a^3+b^3+c^3=3abc$.Therefore, $a^3+b^3+c^3=3abc$.Substitute back the values of $a$, $b$, and $c$: $(x-1)^3+(y-2)^3+(z-3)^3=3(x-1)(y-2)(z-3)$.This matches the standard result for three terms whose sum is zero.Check with $x=1$, $y=2$, $z=3$: then $x+y+z=6$, and both sides equal $0$.Answer:Option D: $3(x-1)(y-2)(z-3)$