1. Given $a^3 + \frac{1}{a^3} = 18$, find $a^2 + \frac{1}{a^2}$. Step 1: Recall the identity for cubes:\ $$a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right)$$ Let $x = a + \frac{1}{a}$. Then, $$x^3 - 3x = 18$$ Step 2: Solve for $x$. $$x^3 - 3x - 18 = 0$$ Try $x=3$: $$3^3 - 3 \times 3 - 18 = 27 - 9 - 18 = 0$$ So, $x = 3$. Step 3: Use $x = a + \frac{1}{a} = 3$ to find $a^2 + \frac{1}{a^2}$. $$\left(a + \frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2}$$ So, $$a^2 + \frac{1}{a^2} = x^2 - 2 = 3^2 - 2 = 9 - 2 = 7$$ --- 2. Given $a^3 + \frac{1}{a^3} = 34\sqrt{5}$, find $a^2 + \frac{1}{a^2}$. Step 1: Using the same substitution $x = a + \frac{1}{a}$ and identity, we have $$x^3 - 3x = 34\sqrt{5}$$ Step 2: Look for $x$ such that $$x^3 - 3x - 34\sqrt{5} = 0$$ Try $x = 2\sqrt{5}$: $$\left(2\sqrt{5}\right)^3 - 3 \times 2\sqrt{5} = 8 \times 5 \sqrt{5} - 6\sqrt{5} = 40\sqrt{5} - 6\sqrt{5} = 34\sqrt{5}$$ So, $x = 2\sqrt{5}$. Step 3: Compute $a^2 + \frac{1}{a^2}$: $$a^2 + \frac{1}{a^2} = x^2 - 2 = (2\sqrt{5})^2 - 2 = 4 \times 5 - 2 = 20 - 2 = 18$$ --- 3. Given $a^3 + \frac{1}{a^3} = 198$, find $a^2 + \frac{1}{a^2}$. Step 1: Using the substitution $x = a + \frac{1}{a}$, we get $$x^3 - 3x = 198$$ Step 2: Solve for $x$: $$x^3 - 3x - 198 = 0$$ Try $x = 6$: $$6^3 - 3 \times 6 = 216 - 18 = 198$$ So, $x = 6$. Step 3: Calculate $a^2 + \frac{1}{a^2}$: $$a^2 + \frac{1}{a^2} = x^2 - 2 = 6^2 - 2 = 36 - 2 = 34$$ Final answers: 1. $7$ 2. $18$ 3. $34$