If $(2p + \frac{1}{p}) = 4$, the value of $(p^3 + \frac{1}{8p^3})$ is

## Concept & Formula This problem tests the manipulation of algebraic identities, specifically adjusting coefficients before applying the cubic expansion formula. 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 the value of $(p^3 + \frac{1}{8p^3})$. Notice that this can be rewritten as a sum of perfect cubes: $(p)^3 + (\frac{1}{2p})^3$. Our base terms are $A = p$ and $B = \frac{1}{2p}$. * **Step 1: Adjust the given equation.** We are given: $2p + \frac{1}{p} = 4$. To match our required base terms $(p + \frac{1}{2p})$, we must divide the entire given equation by 2. $\frac{2p}{2} + \frac{1}{2p} = \frac{4}{2}$ $p + \frac{1}{2p} = 2$ * **Step 2: Cube both sides.** Apply the cubic identity to our new equation: $(p + \frac{1}{2p})^3 = (2)^3$ $(p)^3 + (\frac{1}{2p})^3 + 3(p)(\frac{1}{2p})(p + \frac{1}{2p}) = 8$ * **Step 3: Simplify and solve.** Notice that $p$ and $\frac{1}{2p}$ partially cancel out in the middle term, leaving a coefficient of $\frac{1}{2}$. We also know from Step 1 that $(p + \frac{1}{2p}) = 2$. Substitute these values back in: $p^3 + \frac{1}{8p^3} + 3(\frac{1}{2})(2) = 8$ $p^3 + \frac{1}{8p^3} + 3 = 8$ $p^3 + \frac{1}{8p^3} = 5$ ## Exam Strategy & Shortcut Whenever you see a mismatch between the coefficients in the "given" expression and the "target" expression, your very first step should be to multiply or divide the "given" equation to match the base of the target. Once you aligned $(p + \frac{1}{2p}) = 2$, applying the shortcut formula $k^3 - 3k(ab)$ where $a=1, b=1/2$ yields $2^3 - 3(2)(1/2) = 8 - 3 = 5$ in under 15 seconds. ## Common Pitfall The most frequent mistake is directly cubing the original equation $2p + \frac{1}{p} = 4$ without adjusting it first. This results in terms like $8p^3 + \frac{1}{p^3}$, which is entirely unhelpful for finding the target value, wasting critical exam time. ## Final Answer **Therefore, the correct answer is 5.**