$$(x + \frac{1}{x})(x - \frac{1}{x})(x^2 + \frac{1}{x^2})(x^4 + \frac{1}{x^4})$$is equal to:

Question

Examoogle · Accepted Answer

The correct answer is Option A: $x^8 - \frac{1}{x^8}$. Use the difference of squares identity iteratively: $(a-b)(a+b) = a^2-b^2$. The expression will collapse term-by-term: $(x-1/x)(x+1/x) = x^2-1/x^2$, then $(x^2-1/x^2)(x^2+1/x^2) = x^4-1/x^4$, and finally $(x^4-1/x^4)(x^4+1/x^4) = x^8-1/x^8$.

Step-by-Step Solution & Concepts:
Shortcut Trick: Use the difference of squares identity iteratively: $(a-b)(a+b) = a^2-b^2$. The expression will collapse term-by-term: $(x-1/x)(x+1/x) = x^2-1/x^2$, then $(x^2-1/x^2)(x^2+1/x^2) = x^4-1/x^4$, and finally $(x^4-1/x^4)(x^4+1/x^4) = x^8-1/x^8$.

Formula Used: $$(a - b)(a + b) = a^2 - b^2$$

Given: An algebraic expression: (x + 1/x)(x - 1/x)(x^2 + 1/x^2)(x^4 + 1/x^4)

Steps: Apply identity to the first two terms: Using the algebraic identity $(a - b)(a + b) = a^2 - b^2$, we multiply $(x + \frac{1}{x})$ and $(x - \frac{1}{x})$.. Incorporate the third term: Now multiply the result $(x^2 - \frac{1}{x^2})$ by the next term $(x^2 + \frac{1}{x^2})$ using the same identity.. Incorporate the final term: Finally, multiply the result $(x^4 - \frac{1}{x^4})$ by the last term $(x^4 + \frac{1}{x^4})$.

Answer: A is correct because the successive application of the difference of squares identity results in $x^8 - \frac{1}{x^8}$.

Exam Tip: This iterative collapsing pattern is frequently seen in competitive exams with terms like $(1+x)(1+x^2)(1+x^4)...$; remember to multiply by $(1-x)$ to trigger the telescoping effect.

Examoogle · Answer

Option B: $x^4 - \frac{1}{x^4}$

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Option C: $x^{10} - \frac{1}{x^{10}}$

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Option D: $x^6 - \frac{1}{x^6}$

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