Q.13. Given that

$$\left( a^3 + \frac{2}{b^3} \right) \left( \frac{2}{a^3} - b^3 \right) = \sqrt{2}$$

where $a$ and $b$ are real numbers, what is the least value of $ab$?

Answer Choices

• $-\sqrt{2}$
• $\sqrt{2}$
• $-2\sqrt{2}$
• $2\sqrt{2}$
• $-\frac{\sqrt{2}}{2}$
• $\frac{\sqrt{2}}{2}$
• $-2^{\frac{1}{6}}$
• $2^{\frac{1}{6}}$

Solution

We simplify the given equation by letting $x = a^3$ and $y = b^3$, which reduces the equation to:

$$(x + \frac{2}{y}) \left( \frac{2}{x} - y \right) = \sqrt{2}$$

Expanding and simplifying gives the equation $-xy + \frac{4}{xy} = \sqrt{2}$. Letting $z = xy$, we solve the quadratic equation $z^2 + z\sqrt{2} - 4 = 0$, which gives two possible values for $z$:

$$z = \sqrt{2} \quad \text{or} \quad z = -2\sqrt{2}$$

The least value of ab is