If $(4b^2 + \frac{1}{b^2}) = 2$, then $(8b^3 + \frac{1}{b^3}) = x$

## Concept & Strategy This is an advanced algebraic identity problem because the coefficients of the variable $b$ are not uniform ($4b^2$ vs $\frac{1}{b^2}$). We must carefully identify the base terms before applying squaring and cubing formulas. The core formula required is: $$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$$ ## Step-by-Step Solution * **Analyze the target:** We need to find $(8b^3 + \frac{1}{b^3})$. Notice that this is a perfect cube representation: $(2b)^3 + (\frac{1}{b})^3$. This means our base terms are $A = 2b$ and $B = \frac{1}{b}$. We need to find the value of $(2b + \frac{1}{b})$ to proceed. * **Step 1: Use the given information to find $(2b + \frac{1}{b})$.** Expand the square of our desired base terms: $(2b + \frac{1}{b})^2 = (2b)^2 + (\frac{1}{b})^2 + 2(2b)(\frac{1}{b})$ $(2b + \frac{1}{b})^2 = 4b^2 + \frac{1}{b^2} + 4$ * **Step 2: Substitute the known value.** The question gives us $(4b^2 + \frac{1}{b^2}) = 2$. Substitute this into our expansion: $(2b + \frac{1}{b})^2 = 2 + 4 = 6$ Therefore, $(2b + \frac{1}{b}) = \sqrt{6}$. * **Step 3: Apply the cubic formula.** Cube the base equation: $(2b + \frac{1}{b})^3 = (2b)^3 + (\frac{1}{b})^3 + 3(2b)(\frac{1}{b})(2b + \frac{1}{b})$ Substitute $A = 2b$ and $B = 1/b$: $(\sqrt{6})^3 = (8b^3 + \frac{1}{b^3}) + 3(2)(\sqrt{6})$ * **Step 4: Solve for the target expression.** $6\sqrt{6} = (8b^3 + \frac{1}{b^3}) + 6\sqrt{6}$ Subtract $6\sqrt{6}$ from both sides: $8b^3 + \frac{1}{b^3} = 0$ ## Exam Strategy & Shortcut For any equation mapping $A^3 + B^3$, if you realize that $(A + B)^2$ yields exactly the value of $3AB$, then $A^3 + B^3$ will inherently equal zero. Here, $(2b + 1/b)^2 = 6$, and $3(2b)(1/b) = 6$. Because these match, the terms cancel out during cubing. ## Common Pitfall A severe mistake is treating the $3ab$ term in the cubic formula as just $3$. Because the terms are $2b$ and $1/b$, their product is $2$, meaning the middle multiplier becomes $3 \times 2 = 6$. Forgetting the coefficient of 2 completely derails the calculation. ## Final Answer **Therefore, the correct answer is 0.**