Of squares or cubes?} \\ \text{Write it as a difference of squares.} \end{array} &3xy\left((x^2)^2−(4)2\right) \\ \text{Factor it as a product of conjugates} &3xy(x^2−4)(x^2+4) \\ \text{The first bino...Of squares or cubes?} \\ \text{Write it as a difference of squares.} \end{array} &3xy\left((x^2)^2−(4)2\right) \\ \text{Factor it as a product of conjugates} &3xy(x^2−4)(x^2+4) \\ \text{The first binomial is again a difference of squares.} &3xy\left((x)^2−(2)^2\right)(x^2+4) \\ \text{Factor it as a product of conjugates.} &3xy(x−2)(x+2)(x^2+4) \\ \text{Is the expression factored completely?
Of squares or cubes?} \\ \text{Write it as a difference of squares.} \end{array} &3xy\left((x^2)^2−(4)2\right) \\ \text{Factor it as a product of conjugates} &3xy(x^2−4)(x^2+4) \\ \text{The first bino...Of squares or cubes?} \\ \text{Write it as a difference of squares.} \end{array} &3xy\left((x^2)^2−(4)2\right) \\ \text{Factor it as a product of conjugates} &3xy(x^2−4)(x^2+4) \\ \text{The first binomial is again a difference of squares.} &3xy\left((x)^2−(2)^2\right)(x^2+4) \\ \text{Factor it as a product of conjugates.} &3xy(x−2)(x+2)(x^2+4) \\ \text{Is the expression factored completely?
