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Math Expert V
Joined: 02 Sep 2009
Posts: 65061
If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 77% (01:51) correct 23% (02:20) wrong based on 66 sessions

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If $$x ≠ 1$$ and $$x ≠ -1$$, then $$\frac{(x^3 + 1)(x^3 − 1 )}{(x^2 − 1)}$$ must equal to which of the folllowing?

A. $$x^4 + x^2 + 1$$

B. $$x^4 + x^3 + x + 1$$

C. $$x^6 – 1$$

D. $$x^6 + 1$$

E. $$x^4 + x^2 - 1$$

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Director  D
Joined: 25 Jul 2018
Posts: 731
If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to  [#permalink]

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1
$$\frac{(x^3 + 1)(x^3 − 1 )}{(x^2 − 1)}=\frac{(x^{6}-1)}{ (x^{2}-1)}=$$

= $$\frac{(x^{2}-1)(x^{4}+ x^{2}+1)}{(x^{2}-1)}$$

$$x≠1$$ and $$x≠-1$$
--> $$x^{4}+ x^{2}+1$$

DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1987
Location: India
Re: If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to  [#permalink]

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1
$$x^3+1= (x+1)(x^2-x+1)$$
$$x^3-1= (x-1)(x^2+x+1)$$
$$x^2-1=(x+1)(x-1)$$

Since, $$x^2≠1$$
$$\frac{(x^3 + 1)(x^3 − 1 )}{(x^2 − 1)}$$= $$(x^2-x+1)*(x^2+x+1)= (x^2+1)^2 -x^2= x^4+x^2+1$$

Bunuel wrote:
If $$x ≠ 1$$ and $$x ≠ -1$$, then $$\frac{(x^3 + 1)(x^3 − 1 )}{(x^2 − 1)}$$ must equal to which of the folllowing?

A. $$x^4 + x^2 + 1$$

B. $$x^4 + x^3 + x + 1$$

C. $$x^6 – 1$$

D. $$x^6 + 1$$

E. $$x^4 + x^2 - 1$$

Project PS Butler

Are You Up For the Challenge: 700 Level Questions
GMATWhiz Representative G
Joined: 07 May 2019
Posts: 802
Location: India
Re: If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to  [#permalink]

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1
1
Bunuel wrote:
If $$x ≠ 1$$ and $$x ≠ -1$$, then $$\frac{(x^3 + 1)(x^3 − 1 )}{(x^2 − 1)}$$ must equal to which of the folllowing?

A. $$x^4 + x^2 + 1$$

B. $$x^4 + x^3 + x + 1$$

C. $$x^6 – 1$$

D. $$x^6 + 1$$

E. $$x^4 + x^2 - 1$$

Project PS Butler

Are You Up For the Challenge: 700 Level Questions

Solution:

• $$(x^3 + 1) = (x + 1) (x^2 -x + 1)$$
•$$(x^3 - 1) = (x - 1) (x^2 + x + 1)$$
•$$(x^3 + 1)(x^3 - 1) = (x^2 - 1) [(x^2 + 1)^2 - x ^2] =(x^2 - 1) (x^4 + x^2 + 1)$$
• $$\frac{(x^3 + 1)(x^3 - 1)}{(x^2- 1)} = \frac{(x^2 - 1) (x^4 + x^2 + 1)}{(x^2 - 1)}$$
After cancelling $$(x^2 -1)$$, we get
• $$x^4 + x^2 + 1$$
Hence, the correct answer is Option A.
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Senior Manager  G
Joined: 05 Aug 2019
Posts: 298
Location: India
GMAT 1: 600 Q50 V22 GPA: 4
Re: If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to  [#permalink]

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(x^3+1)(x^3-1)/ (x^2-1)

Numerator x^3-1 = (x-1)(x^2+1+x)
Denominator = (x-1)(x+1)
Numerator x^3+1 = (x+1)(x^2+1-x)

Hence expression comes as
(x^2+x+1)(x^2-x+1)

(x^2+1)^2 -x^2
x^4+x^2+1

Hence A
IESE School Moderator S
Joined: 11 Feb 2019
Posts: 306
Re: If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to  [#permalink]

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I wasn't aware of the formula, so i used substitution method; x=2

Putting x= 2 in given eqn, we get 21.

By putting 2 in all options, only A gives 21.
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NJ Re: If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to   [#permalink] 30 May 2020, 07:45

# If x ≠ 1 and x ≠ -1, then (x^3 + 1)(x^3 − 1 )/(x^2 − 1) must equal to  