(3^3-1)(3^3+1)(3^6+1)(3^12+1)= | GMAT Sprint

Free Practice Question

PS · Problem Solving

555-605

Algebra

$(3^3-1)(3^3+1)(3^6+1)(3^{12}+1)=$

Answer Choices

Correct answer marked below

(3^{24} +1)

(3^{48} -1)

(3^{24} -1)

Correct

3^6(3^{12} -1)

(3^{96} +1)

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Explanation

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Official Explanation

Official

To solve the expression

(3^3-1)(3^3+1)(3^6+1)(3^{12}+1)

, we can use the difference of squares and properties of exponents. First, notice that

(3^3-1)(3^3+1)

can be simplified using the difference of squares:

(3^3-1)(3^3+1) = 3^6 - 1.

Next, we can also simplify

(3^6+1)(3^{12}+1)

using the same principle. We can rewrite

(3^{12}+1)

(3^6)^2 + 1

, which is not directly a difference of squares, but we can use the identity for sums of squares. However, we can also recognize that

(3^6+1)(3^{12}+1)

can be factored as follows:…

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Comments

DarrithFeb 11, 2025

Can you explain more?

KaushikFeb 6, 2025

thanks

LachlanFeb 11, 2025

got this wrong on my practice

pulkitMar 30, 2024

thanks!

JasonJan 25, 2025

thanks

What this question tests

What it tests

Your ability to manipulate linear and quadratic expressions, substitute unknowns, and solve equations under time pressure.

Common trap

Losing a sign when moving terms across the equals sign, or dividing both sides by a variable that could equal zero.

(33−1)(33+1)(36+1)(312+1)=(3^3-1)(3^3+1)(3^6+1)(3^{12}+1)=(33−1)(33+1)(36+1)(312+1)=