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(165)^3 - (163)^3

You can use the formula a^3 - b^3

(165)^2 - (163)^3

Expand (163)^3 as (165 - 2)^3 to simplify further.
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I think I'll have to go to YouTube or something for the ^2 x ^3, but I think I get the ^3 x ^3:

(165)^2 - (164)^2 = (165)^3 - (164)^3 = 329
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Stoneface
(165)^2 - (164)^2 =

A. 1
B. 100
C. 229
D. 329
E. 349

Difference between square of two consecutive numbers is = the sum of the of the two consecutive numbers

165+164 = 329

D
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Stoneface
(165)^2 - (164)^2 =

A. 1
B. 100
C. 229
D. 329
E. 349


Using the formula:

(a+1)^2 - a^2 = 2a+1

So, Answer = 164*2 + 1
= 328+1 = 329 = Answer = D
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Where can these rules for this particular problem be found?
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GMATcrusher
Where can these rules for this particular problem be found?

See Bunuel's signature for the Math exponents resources link.
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Stoneface
I think I'll have to go to YouTube or something for the ^2 x ^3, but I think I get the ^3 x ^3:

(165)^2 - (164)^2 = (165)^3 - (164)^3 = 329

No, the expansion is different for both, please see the exponents theory in the Math chapter resources section.

(165)^2 - (164)^2 = 329

(165)^3 - (164)^3 = 81181
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GMATcrusher
Where can these rules for this particular problem be found?

Check Tips and hints on Algebra here: algebra-tips-and-hints-175003.html

Also, I suggest you to go through the following post ALL YOU NEED FOR QUANT and check links in my signature below.

Hope it helps.
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I solved it slightly different:

165^2-(165-1)^2= 165^2-(165^2+1^2-2*165*1)= 165-165-1+330= 329
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Stoneface
(165)^2 - (164)^2 =

A. 1
B. 100
C. 229
D. 329
E. 349

\((165)^2 - (164)^2\)

\((165 + 164) (165 - 164)\)

\(329 * 1\)

(D)
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Stoneface
(165)^2 - (164)^2 =

A. 1
B. 100
C. 229
D. 329
E. 349

Solution:

Recall that the difference of squares (a^2 - b^2) factors as (a + b)(a - b). Using this fact, we have:

(165)^2 - (164)^2 = (165 + 164)(165 - 164) = 329 x 1 = 329

Answer: D
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