1. Problem 66: Given $4a^2 + 9b^2 = 15$ and $ab = 1$, find $(2a + 3b)^2$. Step 1: Expand $(2a + 3b)^2 = 4a^2 + 12ab + 9b^2$. Step 2: Substitute known values: $4a^2 + 9b^2 = 15$ and $ab = 1$ into the expansion. Step 3: Thus, $(2a + 3b)^2 = 15 + 12 \times 1 = 15 + 12 = 27$. 2. Problem 67: Given $x - 3y = 12$ and $xy = 6$, find $x^2 + 9y^2$. Step 1: Recognize that $x^2 + 9y^2 = (x - 3y)^2 + 2 \cdot x \cdot (-3y)$ (since $(x - 3y)^2 = x^2 - 2 \cdot 3xy + 9y^2$). Step 2: Rearranged, $x^2 + 9y^2 = (x - 3y)^2 + 6xy$. Step 3: Substitute $x - 3y = 12$ and $xy=6$: $$x^2 + 9y^2 = 12^2 + 6 \cdot 6 = 144 + 36 = 180$$. 3. Problem 68: Given $2x + 3y = 10$ and $xy = 5$, find $4x^2 + 9y^2$. Step 1: Expand $(2x + 3y)^2 = 4x^2 + 12xy + 9y^2$. Step 2: Rearranged, $4x^2 + 9y^2 = (2x + 3y)^2 - 12xy$. Step 3: Substitute known values: $(2x + 3y) = 10$ and $xy=5$: $$4x^2 + 9y^2 = 10^2 - 12 \times 5 = 100 - 60 = 40$$. 4. Problem 69: Given $x^2 + \frac{9}{x^2} = 10$, find $(x + \frac{3}{x})^2$. Step 1: Expand $(x + \frac{3}{x})^2 = x^2 + 2 \cdot x \cdot \frac{3}{x} + \frac{9}{x^2} = x^2 + 6 + \frac{9}{x^2}$. Step 2: Note that $x^2 + \frac{9}{x^2} = 10$ (given). Step 3: Substitute: $$(x + \frac{3}{x})^2 = 10 + 6 = 16$$. 5. Problem 70: Given $a^2 + \frac{4}{a^2} = 6$, find $(a + \frac{2}{a})^2$. Step 1: Expand $(a + \frac{2}{a})^2 = a^2 + 2 \cdot a \cdot \frac{2}{a} + \frac{4}{a^2} = a^2 + 4 + \frac{4}{a^2}$. Step 2: Given $a^2 + \frac{4}{a^2} = 6$. Step 3: Substitute: $$(a + \frac{2}{a})^2 = 6 + 4 = 10$$.