Question
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2. \( \left[(p+4)\left(p^{2}-4 p+16\right)\right]^{2} \) \( =\left[p^{3}+64\right]^{2} \) \( = \) \( p^{3}+128 p^{3}+4096 \)

Ask by Mcguire Davies. in South Africa
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

\[ \left[(p + 4)(p^{2} - 4p + 16)\right]^{2} = p^{6} + 128p^{3} + 4096 \]

Solution

You're on the right track with expanding \(\left[(p + 4)(p^{2} - 4p + 16)\right]^{2}\). However, there's a small mistake in the final expansion step. Let's go through it step by step to clarify: 1. **Simplify the Inner Expression:** \[ (p + 4)(p^{2} - 4p + 16) = p^{3} + 4p^{2} - 4p^{2} - 16p + 16p + 64 = p^{3} + 64 \] So, the expression becomes: \[ \left[(p + 4)(p^{2} - 4p + 16)\right]^{2} = (p^{3} + 64)^{2} \] 2. **Expand \((p^{3} + 64)^{2}\):** Use the formula \((a + b)^2 = a^2 + 2ab + b^2\), where \(a = p^{3}\) and \(b = 64\): \[ (p^{3} + 64)^{2} = (p^{3})^{2} + 2 \cdot p^{3} \cdot 64 + (64)^{2} = p^{6} + 128p^{3} + 4096 \] 3. **Final Result:** \[ \left[(p + 4)(p^{2} - 4p + 16)\right]^{2} = p^{6} + 128p^{3} + 4096 \] **Summary:** \[ \left[(p + 4)(p^{2} - 4p + 16)\right]^{2} = p^{6} + 128p^{3} + 4096 \] This corrects the expansion by including the \(p^{6}\) term, which was previously omitted.

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Mind Expander

Let’s clear things up here! The expression \(\left[(p+4)(p^{2}-4p+16)\right]^{2}\) actually simplifies differently. When we multiply \((p+4)\) by \((p^{2}-4p+16)\), we end up with a polynomial that, when squared, gives a different result than just \(\left[p^{3}+64\right]^{2}\). To evaluate \(\left[p^{3}+64\right]^{2}\), we apply the binomial expansion: \[ (p^3 + 64)^2 = p^6 + 128p^3 + 4096 \] So make sure to double-check your steps when expanding or simplifying expressions! Once you understand the binomial theorem, you can see how it applies to not just numbers but any polynomial form—making it easier to manage algebraic expressions! Plus, getting comfortable with these expansions paves the way for dealing with more complex polynomials down the line.

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