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# X^8 - Y^8 =

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Re: X^8 - Y^8 = [#permalink]
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BM wrote:
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. $$(X^2 - Y^2)^4$$

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, $$X^8 - Y^8 =(X^4 + Y^4)(X^4 - Y^4)$$

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...
$$(X^4 + Y^4)(X^2 + Y^2)(X^2 - Y^2)=(X^4 + Y^4)(X^2 + Y^2)(X+Y)(X-Y)$$

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Looks like 500-level question))
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ulm wrote:
Looks like 500-level question))

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!
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I go with option B

It is definitely not a 700 level problem
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Re: X^8 - Y^8 = [#permalink]
BM wrote:
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. $$(X^2 - Y^2)^4$$

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
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Re: X^8 - Y^8 = [#permalink]
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BM wrote:
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. $$(X^2 - Y^2)^4$$

Bunuel could you please explain this?
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Re: X^8 - Y^8 = [#permalink]
shivamtibrewala wrote:
Bunuel could you please explain this?

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?
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Re: X^8 - Y^8 = [#permalink]
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This question can be solved by taking the difference of squares multiple times.

$$x^8-y^8$$

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

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

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

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Re: X^8 - Y^8 = [#permalink]
BM wrote:
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. $$(X^2 - Y^2)^4$$

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)

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Re: X^8 - Y^8 = [#permalink]
$$X^8 - Y^8 = (X^4+Y^4)(X^4-Y^4) = (X^4+Y^4)(X^2+Y^2)(X^2-Y^2) = \\ (X^4+Y^4)(X^2+Y^2)(X+Y)(X-Y)$$

IMO B

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Re: X^8 - Y^8 = [#permalink]
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Re: X^8 - Y^8 = [#permalink]
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