Answer:
[tex](3\, y + 2\, z)^{2} - (3\, y - 2\, z)^{2} = 24\, y\, z[/tex]
Step-by-step explanation:
Make use of the fact that for any [tex]a[/tex] and [tex]b[/tex]:
[tex]\begin{aligned}& (a + b)\, (a - b) \\ =\; & a^{2} - a\, b + a\, b - b^{2} \\ =\; & a^{2} - b^{2}\end{aligned}[/tex].
In other words, the difference [tex](a^{2} - b^{2})[/tex] between two squares could be written as the product of [tex](a + b)[/tex] and [tex](a - b)[/tex].
Apply this identity to rewrite and simplify the expression in this question. In this example, [tex]a = 3\, y + 2\, z[/tex] whereas [tex]b = 3\, y - 2\, z[/tex].
[tex]\begin{aligned} & \underbrace{(3\, y + 2\, z)^{2}}_{a^{2}} - \underbrace{(3\, y - 2\, z)^{2}}_{b^{2}} \\ =\; & \underbrace{((3\, y + 2\, z) + (3\, y - 2\, z))}_{(a + b)}\\ &\times \underbrace{((3\, y + 2\, z) - (3\, y - 2\, z))}_{(a - b)} \\ =\; &6\, y\, (3\, y + 2\, z - 3\, y + 2\, z) \\ =\; & 24\, y \, z \end{aligned}[/tex].
