Question

find the binomial that completes the factorization.
$3x^{3}y - 3y = 3y(\quad)(x^{2} + x + 1)$

Explanation:

Step1: Factor out common term

$3x^3y - 3y = 3y(x^3 - 1)$

Step2: Apply difference of cubes

Recall $a^3 - b^3=(a-b)(a^2+ab+b^2)$. Here $a=x$, $b=1$, so $x^3 - 1=(x-1)(x^2+x+1)$

Step3: Match to given factorization

Substitute back: $3y(x^3 - 1)=3y(x-1)(x^2+x+1)$

Answer:

$x-1$