To factorise the expression $3x^3 - 12xy^2$ completely, we will find common factors and apply algebraic identities. Step 1: Identify the greatest common factor (GCF). The terms are $3x^3$ and $-12xy^2$. The numerical coefficients are 3 and -12. The GCF of 3 and 12 is 3. The variable parts are $x^3$ and $x$. The GCF of $x^3$ and $x$ is $x$. There is no common factor for $y^2$ as it only appears in the second term. So, the GCF of the entire expression is $3x$. Step 2: Factor out the GCF. Divide each term by $3x$: $\frac{3x^3}{3x} = x^2$ $\frac{-12xy^2}{3x} = -4y^2$ So, the expression becomes $3x(x^2 - 4y^2)$. Step 3: Apply the difference of squares identity. The term inside the parentheses, $x^2 - 4y^2$, is in the form of $a^2 - b^2$. Here, $a = x$ and $b = \sqrt{4y^2} = 2y$. The difference of squares identity is $a^2 - b^2 = (a-b)(a+b)$. So, $x^2 - 4y^2 = (x - 2y)(x + 2y)$. Step 4: Combine all factors. Substitute the factored form of $x^2 - 4y^2$ back into the expression from Step 2. $$3x(x^2 - 4y^2) = 3x(x - 2y)(x + 2y)$$ The completely factorised expression is $3x(x - 2y)(x + 2y)$. \boxed{3x(x - 2y)(x + 2y)} That's 2 down. 3 left today — send the next one.