If $$a$$ is positive and $$a^2 + \frac{1}{a^2} = 7,$$ then $$a^3 + \frac{1}{a^3} = ?$$

$$a^2 + \frac{1}{a^2} = 7$$

Addition 2 in both sides of equation.

$$a^2 + \frac{1}{a^2} + 2 = 7+2$$

$$a^2 + \frac{1}{a^2} + 2 = 9$$    Eq.(1)

Eq.(1) is making the formula of $$(a+\frac{1}{a})^{2}$$.

After removing the square got $$(a+\frac{1}{a}) = \pm 3$$

In question, it is mentioned that value of a is positive.

So $$(a+\frac{1}{a}) = 3$$    Eq.(2)

In Eq.(2) apply formula $$(a+\frac{1}{a})^{3}$$.

So $$(a+\frac{1}{a})^{3} = a^{3} + (\frac{1}{a})^{3} + 3 \times a \times (\frac{1}{a}) [a + \frac{1}{a}]$$

$$(a+\frac{1}{a})^{3} = a^{3} + (\frac{1}{a})^{3} + 3[a + \frac{1}{a}]$$    Eq.(3)

Put Eq.(2) in Eq.(3).

$$(3)^{3} = a^{3} + (\frac{1}{a})^{3} + 3\times3$$

$$27 = a^{3} + (\frac{1}{a})^{3} + 9$$

$$27-9 = a^{3} + (\frac{1}{a})^{3}$$

$$18 = a^{3} + (\frac{1}{a})^{3}$$

$$a^{3} + (\frac{1}{a})^{3} = 18$$