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# If x is not equal to zero, and x+1/x = 3

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Intern
Joined: 01 May 2017
Posts: 18
Location: United States (IL)
Concentration: Technology, Strategy
GPA: 3.95
If x is not equal to zero, and x+1/x = 3  [#permalink]

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18 Mar 2018, 12:13
3
00:00

Difficulty:

75% (hard)

Question Stats:

59% (02:08) correct 41% (03:01) wrong based on 63 sessions

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If x is not equal to zero, and x+1/x = 3, then what is the value of x^6 + (1/x)^6?

A. 320
B. 322
C. 324
D. 326
E. 328
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3197
Location: India
GPA: 3.12
If x is not equal to zero, and x+1/x = 3  [#permalink]

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18 Mar 2018, 12:52
2
If x is not equal to zero, and x+1/x = 3, then what is the value of x^6 + (1/x)^6?

A. 320
B. 322
C. 324
D. 326
E. 328

Given: ($$x + \frac{1}{x}$$) = 3

Squaring on both sides, we get $$(x + \frac{1}{x})^2 = 3^2$$ -> $$x^2 +(\frac{1}{x^2}) + 2 = 9$$ -> $$x^2 +(\frac{1}{x^2}) = 7$$

Applying cube on both sides, we get $$(x^2 +(\frac{1}{x^2}))^3 = 7^3$$

Formula used: $$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$$

The equation simplifies to $$(x^2)^3 + (\frac{1}{x^2})^3 + 3*x^2*\frac{1}{x^2}(x^2 +(\frac{1}{x^2})) = 343$$

Therefore, the value of $$x^6 + (\frac{1}{x^6}) + 3*7 = 343$$ -> $$x^6 + (\frac{1}{x^6}) = 343 - 21 = 322$$(Option B)
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Math Expert
Joined: 02 Sep 2009
Posts: 50042
Re: If x is not equal to zero, and x+1/x = 3  [#permalink]

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19 Mar 2018, 10:28
2
If x is not equal to zero, and x+1/x = 3, then what is the value of x^6 + (1/x)^6?

A. 320
B. 322
C. 324
D. 326
E. 328

Similar question to practice: https://gmatclub.com/forum/if-x-is-not- ... 26295.html
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Intern
Joined: 30 Nov 2016
Posts: 35
GMAT 1: 650 Q47 V33
Re: If x is not equal to zero, and x+1/x = 3  [#permalink]

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03 Apr 2018, 03:00
My approach(I found it easy with squaring and proceeding):-
Given: x+1/x = 3
=> x^2 + (1/x)^2 + 2*x*(1/x) = 9
=> x^2 + (1/x)^2 = 7......(1)

Now, square eqn 1:-
x^4 + (1/x)^4 = 47.....(2)

Now multiply eqn 1 & 2:-
=> x^6 + (1/x)^2 + x^2 + (1/x)^6 = 329;
=> x^6 + 7 + (1/x)^6 = 329.....got "7" using eqn 1
=> x^6 + (1/x)^6 = 322

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Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4095
Location: India
GPA: 3.5
Re: If x is not equal to zero, and x+1/x = 3  [#permalink]

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03 Apr 2018, 09:01
chandra004 wrote:
My approach(I found it easy with squaring and proceeding):-
Given: x+1/x = 3
=> x^2 + (1/x)^2 + 2*x*(1/x) = 9
=> x^2 + (1/x)^2 = 7......(1)

Now, square eqn 1:-
x^4 + (1/x)^4 = 47.....(2)

Now multiply eqn 1 & 2:-
=> x^6 + (1/x)^2 + x^2 + (1/x)^6 = 329;
=> x^6 + 7 + (1/x)^6 = 329.....got "7" using eqn 1
=> x^6 + (1/x)^6 = 322

Possible, but pushpitkc has shown the fastest possible approach for exam hall situation, it is best to adopt that method.
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Abhishek....

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Re: If x is not equal to zero, and x+1/x = 3 &nbs [#permalink] 03 Apr 2018, 09:01
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