Manhattan GMAT Challenge Problem of the Week - 26 Feb 10

by , Feb 26, 2010

Welcome back to this week's Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 minutes and go!

Question

[pmath]x^8 - y^8[/pmath] =

(A) [pmath](x^4 - y^4)^2[/pmath]

(B) [pmath](x^4 + y^4)(x^2 + y^2)(x + y)(x - y)[/pmath]

(C) [pmath](x^6 + y^2)(x^2 - y^6)[/pmath]

(D) [pmath](x^4 - y^4)(x^2 - y^2)(x - y)(x + y)[/pmath]

(E) [pmath](x^2 - y^2)^4[/pmath]

Solution

You need to recognize the expression as a difference of squares. Like all other even powers, [pmath]x^8[/pmath] is a square, equal to [pmath](x^4)^2[/pmath], so we use the template [pmath]a^2 - b^2[/pmath] = (a + b)(a b), with [pmath]a = x^4[/pmath] and [pmath]b =y^4[/pmath]:

[pmath]x^8 - y^8 = (x^4)^2 - (y^4)^2 = (x^4 + y^4)(x^4 - y^4)[/pmath]

We continue breaking down the second part of the resulting expression, which is also a difference of squares.

[pmath]x^8 - y^8 = (x^4)^2 - (y^4)^2 = (x^4 + y^4)(x^2 + y^2)(x^2 - y^2)[/pmath]

And were not done yet, because the last expression is of course also a difference of squares!

[pmath]x^8 - y^8 = [/pmath]

[pmath](x^4)^2 - (y^4)^2 =[/pmath]

[pmath](x^4 + y^4)(x^2 + y^2)(x^2 - y^2) =[/pmath]

[pmath](x^4 + y^4)(x^2 + y^2)(x + y)(x - y)[/pmath]

This final product matches the expression in choice (B), so the correct answer is (B).

Plugging numbers is probably too time-consuming in this case. Among positive integers, only 0 and 1 are easy to compute the eighth power of (unless you've memorized that 28 = 256). Moreover, several of the answer choices are designed to give you 0 if you choose x = y = 1.

If you did plug in x = 2 and y = 1, then you would get the following for choice (B):

256 1 = 255 = (16 + 1)(4 + 1)(2 + 1)(2 1) = (17)(5)(3).

If you happen to know already that 28 = 256, then you could get 255 as your target number relatively quickly. Since 255 ends in 5, it must be divisible by 5. No choice besides (B) gives you 5 as a factor if you plug in x = 2 and y = 1, so you wouldnt need to compute the actual values of every choice. However, its still the case that the best way to do this problem is to recognize the original expression as a difference of squares, and then factor.

Again, the correct answer is (B) [pmath](x^4 + y^4)(x^2 + y^2)(x + y)(x - y)[/pmath].

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