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If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ?

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Bunuel
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If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ? [#permalink] If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ?   12 Aug 2021, 03:30
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If \(x > 1\) and \(x^2 + \frac{1}{x^2} = 83\), then what is the value of \(x^3 - \frac{1}{x^3}\) ?

(A) 744
(B) 750
(C) 756
(D) 760
(E) 764
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C
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chetan2u
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Re: If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ? [#permalink] If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ?   12 Aug 2021, 03:49
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Bunuel wrote:
If \(x > 1\) and \(x^2 + \frac{1}{x^2} = 83\), then what is the value of \(x^3 - \frac{1}{x^3}\) ?

(A) 744
(B) 750
(C) 756
(D) 760
(E) 764



\(x^2 + \frac{1}{x^2} = 83\)

\(x^2 + \frac{1}{x^2} -2*x*\frac{1}{x} = 83-2=81\)

\((x-\frac{1}{x})^2=81\)

Taking square root on both sides,

\((x-\frac{1}{x})=9=-9\)

Taking cube, we get

\((x-\frac{1}{x})^3=9^3=729\)

\(x^3-\frac{1}{x^3}-3*x*\frac{1}{x}(x-\frac{1}{x})=729\)

\(x^3-\frac{1}{x^3}-3(9)=729\)

\(x^3-\frac{1}{x^3}=729+27=756\)


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If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ? [#permalink] If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ?   12 Aug 2021, 04:21
\(x^2 + 1/x^2\) = 83 ..................(1)

=> \(x^2 + 1/x^2\) - 2 = 81

=> \((x - 1/x)^2 = 9^2\)

=> \(x - 1/x = 9\) .....................(2)

Now \(x^3 - 1/x^3 = (x - 1/x)(x^2 + 1/x^2 + 1)\) = 9 * 84 = 756


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Re: If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ? [#permalink] If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ?   28 Oct 2021, 13:40
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chetan2u wrote:
Bunuel wrote:
If \(x > 1\) and \(x^2 + \frac{1}{x^2} = 83\), then what is the value of \(x^3 - \frac{1}{x^3}\) ?

(A) 744
(B) 750
(C) 756
(D) 760
(E) 764



\(x^2 + \frac{1}{x^2} = 83\)

\(x^2 + \frac{1}{x^2} -2*x*\frac{1}{x} = 83-2=81\)

\((x-\frac{1}{x})^2=81\)

Taking square root on both sides,

\((x-\frac{1}{x})=9=-9\)

Taking cube, we get

\((x-\frac{1}{x})^3=9^3=729\)

\(x^3-\frac{1}{x^3}-3*x*\frac{1}{x}(x-\frac{1}{x})=729\)

\(x^3-\frac{1}{x^3}-3(9)=729\)

\(x^3-\frac{1}{x^3}=729+27=756\)


C

Hi Chetan2u,
Would you please help to understand how you got 9 from (x−1/x) ...
I got everything up to that point. x3−1x3−3∗x∗1x(x−1/x)=729

thank you
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Re: If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ? [#permalink] If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ?   28 Oct 2021, 17:42
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jdcsantos wrote:
chetan2u wrote:
Bunuel wrote:
If \(x > 1\) and \(x^2 + \frac{1}{x^2} = 83\), then what is the value of \(x^3 - \frac{1}{x^3}\) ?

(A) 744
(B) 750
(C) 756
(D) 760
(E) 764



\(x^2 + \frac{1}{x^2} = 83\)

\(x^2 + \frac{1}{x^2} -2*x*\frac{1}{x} = 83-2=81\)

\((x-\frac{1}{x})^2=81\)

Taking square root on both sides,

\((x-\frac{1}{x})=9=-9\)

Taking cube, we get

\((x-\frac{1}{x})^3=9^3=729\)

\(x^3-\frac{1}{x^3}-3*x*\frac{1}{x}(x-\frac{1}{x})=729\)

\(x^3-\frac{1}{x^3}-3(9)=729\)

\(x^3-\frac{1}{x^3}=729+27=756\)


C

Hi Chetan2u,
Would you please help to understand how you got 9 from (x−1/x) ...
I got everything up to that point. x3−1x3−3∗x∗1x(x−1/x)=729

thank you


\(x^3-\frac{1}{x^3}-3*x*\frac{1}{x}(x-\frac{1}{x})=729\)

\(x^3-\frac{1}{x^3}-3(x-\frac{1}{x})=729\)

Now \(x-\frac{1}{x}=9, so substitute this value. \)

\(x^3-\frac{1}{x^3}-3*9=729\)
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Re: If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ? [#permalink] If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ?   27 May 2023, 02:08
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Re: If x > 1 and x^2 + 1/x^2 = 83, then what is the value of x^3 - 1/x^3 ? [#permalink]
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