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Let
A = 1 + sqrt(2) + sqrt(3) and B = sqrt(2) + sqrt(3).

=> A^2 − B^2 = (A − B)(A + B)
= (1 + sqrt(2) + sqrt(3) − sqrt(2) − sqrt(3)) × (1 + sqrt(2) + sqrt(3) + sqrt(2) + sqrt(3))
= 1 × (1 + 2sqrt(2) + 2sqrt(3))
= 1 + 2sqrt(2) + 2sqrt(3).
Option: D
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Think of sqrt 2 + sqrt 3 as x

(1 + x)^2 - x^2 = (1 + 2x + x^2) - x^2= 1 + 2x
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Let x be \(\sqrt{2}+ \sqrt{3}\)

We can rewrite the expression as:

\((1+x)^2-x^2=\)

\(=x^2+1+2x-x^2=\)

\(=1 + 2x=\)

\(=1 + 2( \sqrt{2}+ \sqrt{3})=\)

\(=1 + 2\sqrt{2}+ 2\sqrt{3})\)


Answer: D
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Which of the following is equal to \((1 + \sqrt{2} + \sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^2\)?

This question could be done via expansion of the two squared expressions. At the same time, there's a much faster way: using the difference of squares pattern that commonly appears in GMAT Problem Solving questions.

Considering \((1 + \sqrt{2} + \sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^2\) carefully, we can see that it's a difference of squares, in the form \(a^2 - b^2\).

Accordingly, we can use the fact that \(a^2 - b^2 = (a + b)(a - b)\) to answer this question quickly.

\((1 + \sqrt{2} + \sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^2\)

\(((1 + \sqrt{2} + \sqrt{3}) + (\sqrt{2}+\sqrt{3}))((1 + \sqrt{2} + \sqrt{3}) - (\sqrt{2}+\sqrt{3}))\)

\((1 + 2\sqrt{2} + 2\sqrt{3})(1)\)

\(1 + 2\sqrt{2} + 2\sqrt{3}\)

A. \(1\)

B. \(2\)

C. \(-1 + 2\sqrt{2}\)

D. \(1 + 2\sqrt{2} + 2\sqrt{3}\)

E. \(2 + 2\sqrt{2} + 2\sqrt{3}\)


Correct answer: D
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placeatex
This is a question from official mock #3, and the first time I have come across (a+b+c)^2 being tested in the GMAT during my (albeit fairly brief) studies

placeatex
It is actually testing difference of squares: \(a^2 - b^2 = (a+b)(a-b)\) and that is tested very often in GMAT. You do not need to use \((a+b+c)^2\) here and if you did, it would take too much time. It is a bit cumbersome for GMAT though I still recommend memorizing it (in case nothing else comes to mind). Do note though that you can simply do the multiplication (a+b+c)(a+b+c) to get the answer but chances of a making a careless calculation mistake increase exponentially in that case.
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