X^8 - Y^8 =

Question

X^8 - Y^8 =

A.  (X^4 - Y^4)^2 
B.  (X^4 + Y^4)(X^2 + Y^2)(X + Y)(X - Y) 
C.  (X^6 + Y^2)(X^2 - Y^6) 
D.  (X^4 - Y^4)(X^2 - Y^2)(X - Y)(X + Y) 
E.  (X^2 - Y^2)^4

whiplash2411 · Accepted Answer

The solution is just a basic application of the principle that  x^2 - y^2 = (x+y)(x-y)

So, when we simplify, we treat  x^8 = (x^4)^2 = ((x^2)^2)^2

So we get the following:

x^8 - y^8 = (x^4+y^4)(x^4-y^4) = (x^4+y^4)(x^2+y^2)(x^2 - y^2) =(x^4+y^4)(x^2+y^2)(x+y)(x-y)

So, the answer is B.

tkarthi4u · Answer

Apply a^2 - b^2 = (a-b)(a+b)

(x^4)^2 - (y^4)^2

( x^4 + y^4) ( x^4 - y^4)

( x^4 + y^4) (x^2+y^2)(x^2-y^2)

( x^4 + y^4) (x^2+y^2)(x+y)(x-y)

So B

BrentGMATPrepNow · Answer

Since  (X^4)^2 = X^8 , and  (Y^4)^2 = Y^8 , we can see that  X^8 - Y^8  is a difference of squares

So,  X^8 - Y^8 =(X^4 + Y^4)(X^4 - Y^4) 
Scan the answer choices....not there!!

Aha, now we must recognize that  X^4 - Y^4  is a difference of squares, which we can factor to get:
 (X^4 + Y^4)(X^4 - Y^4) = (X^4 + Y^4)(X^2 + Y^2)(X^2 - Y^2)

Since  X^2 - Y^2  is another difference of squares, we can continue factoring to get...
 (X^4 + Y^4)(X^2 + Y^2)(X^2 - Y^2)=(X^4 + Y^4)(X^2 + Y^2)(X+Y)(X-Y)

Answer: B

ulm · Answer

Looks like 500-level question))

BM · Answer

Agreed. I personally think that it belongs in the 600-level category, but MGMAT considers it a 700-level question. I just wanted to stay as faithful to the question source as possible. I'll post a harder one to make up for it! :wink:

smartmundu · Answer

I go with option B

It is definitely not a 700 level problem

PareshGmat · Answer

I solved this is hardly 3 seconds, would tell how?

x^8 - y^8 It has even power( 8 = 2^3 ) & -ve sign in between, so fully expandable to the least power of x & y i.e (x+y) All +ve powers cannot be expanded, so out of the 5 options, just search for (x-y) & terms with high +ve powers Only option B fits in

shivamtibrewala · Answer

Bunuel  could you please explain this?

abhimahna · Answer

Hey  shivamtibrewala  ,

It is the simple use of formula  a^2 - b^2 = (a+b)(a-b)

Consider  a = x^4  and  b = y^4  first and expand as mentioned above.

Once done you will get  (x^4 +y^4)(x^4 - y^4)

Now, this is again the usage of same formula. Here consider  a = x^2  and  b = y^2

Once done you will get  (x^2+y^2)(x^2 - y^2)

Same goes for  (x^2 - y^2) .

hence you will get :

x^8 - y^8 = (x^4+y^4)(x^4-y^4) = (x^4+y^4)(x^2+y^2)(x^2 - y^2) =(x^4+y^4)(x^2+y^2)(x+y)(x-y)

Does that make sense?

CatchTwentyTwo · Answer

This question can be solved by taking the difference of squares multiple times.

x^8-y^8

(x^4+y^4)(x^4-y^4)

(x^4+y^4)(x^2+y^2)(x^2-y^2)

(x^4+y^4)(x^2+y^2)(x+y)(x-y)

Answer B.

ScottTargetTestPrep · Answer

X^8 - Y^8 =

(X^4 + Y^4)(X^4 - Y^4)

Notice that the second expression is a difference of fourth powers, so it can be factored further as:

(X^4 + Y^4)(X^2 - Y^2)(X^2 + Y^2)

The middle expression is a difference of squares, so it can be factored further as:
(X^4 + Y^4)(X - Y)(X + Y)(X^2 + Y^2)

Answer: B

Kinshook · Answer

X^8 - Y^8 = (X^4+Y^4)(X^4-Y^4) = (X^4+Y^4)(X^2+Y^2)(X^2-Y^2) = 
(X^4+Y^4)(X^2+Y^2)(X+Y)(X-Y)

IMO B

Posted from my mobile device

bumpbot · Answer

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X^8 - Y^8 =

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