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 August 9th, 2018, 10:23 AM #1 Member   Joined: Oct 2012 Posts: 78 Thanks: 0 nth roots If (r)^(-1/4) - (r)^(1/4) = 14. Find the value of (r)^(-1/6) + (r)^(1/6)
 August 9th, 2018, 12:05 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 r = 19601 - 13860*sqrt(2) Thanks from topsquark
 August 9th, 2018, 12:23 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 The answer to this thing is 6. It can be done via brute force using hyperbolic trig multi-angle formulas but there is probably a very clever way lurking in here that someone like Idea or SlipEternal will see. Thanks from topsquark
August 9th, 2018, 02:01 PM   #4
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Quote:
 Originally Posted by fahad nasir If (r)^(-1/4) - (r)^(1/4) = 14. Find the value of (r)^(-1/6) + (r)^(1/6)
I'd try a u-substitution.

$u = r^{1/4} \text { and } 14 = r^{-1/4} - r^{1/4} = u^{-1} - u = \dfrac{1}{u} - u \implies$

$\dfrac{1 - u^2}{u} = 14 \implies u^2 + 14u - 1 = 0 \implies u = \dfrac{-\ 14 \pm \sqrt{14^2 - 4(1)(-\ 1)}}{2} = -\ 7 \pm 5 \sqrt{2}.$

Is that correct?

$\dfrac{1}{-\ 7 - 5 \sqrt{2}} - (-\ 7 - 5\sqrt{2}) = \dfrac{-\ 1}{7 + 5 \sqrt{2}} + 7 + 5 \sqrt{2} = \dfrac{49 + 2 * 7 * 5 \sqrt{2} + 25 * 2 - 1}{7 + 5 \sqrt{2}} =$

$\dfrac{98 + 70\sqrt{2}}{7 + 5 \sqrt{2}} = \dfrac{14(7 + 5 \sqrt{2}}{7 + 5 \sqrt{2}} = 14.$

$\dfrac{1}{-\ 7 + 5 \sqrt{2}} - (-\ 7 + 5\sqrt{2}) = \dfrac{-\ 1}{7 - 5 \sqrt{2}} + 7 - 5 \sqrt{2} = \dfrac{49 - 2 * 7 * 5 \sqrt{2} + 25 * 2 - 1}{7 + 5 \sqrt{2}} =$

$\dfrac{98 - 70\sqrt{2}}{7 - 5 \sqrt{2}} = \dfrac{14(7 - 5 \sqrt{2}}{7 + 5 \sqrt{2}} = 14.$

Looks OK.

$v = r^{1/6} \text { and } u = r^{1/4} \implies v^6 = r = u^4 \implies v = (u^4)^{1/6} = u^{2/3}.$

$\therefore r^{-1/6} - r^{1/6} = v^{-1} - v = \dfrac{1}{v} - v = \dfrac{1 - v^2}{v} =$

$\dfrac{1 - (u^{2/3})^2}{u^{2/3}} = \dfrac{1 - \sqrt[3]{u^4}}{\sqrt[3]{u^2}}.$

$u = -\ 7 \pm 5 \sqrt{2} \implies u^2 = 99 \pm 70 \sqrt{2} \implies u^4 = 19601 \pm 13860 \sqrt{2}.$

$\therefore r^{-1/6} - r^{1/6} = \dfrac{1 - \sqrt[3]{19601 \pm 13860 \sqrt{2}}}{\sqrt[3]{99 \pm 70 \sqrt{2}}}.$

Please check this because it was very ugly.

 August 9th, 2018, 03:34 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Let: $\displaystyle u=r^{\Large\frac{1}{12}}$ And so: $\displaystyle u^{-3}-u^{3}=14$ Let: $\displaystyle S=u^{-1}-u^{1}$ Thus: $\displaystyle S^2=u^{-2}-2+u^{2}$ And: $\displaystyle S^3=u^{-3}-3u^{-1}+3u-u^{3}$ This implies: $\displaystyle S^3+3S-14=0$ Or: $\displaystyle (S-2)\left(S^2+2S+7\right)=0$ Assuming $\displaystyle S$ is real, we then find: $\displaystyle S=2$ Thus: $\displaystyle S^2+2=u^{-2}+u^{2}=r^{-\Large\frac{1}{6}}+r^{\Large\frac{1}{6}}=6$ Thanks from greg1313, Maschke, fahad nasir and 2 others
August 9th, 2018, 09:15 PM   #6
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Quote:
 Originally Posted by JeffM1 $r^{-1/6} - r^{1/6} =$ . . . Please check this
It's $r^{1/6} + r^{1/6}$ that is to be calculated.

$r = (\pm\sqrt2 - 1)^{12}$, from which the answer of 6 is easily obtained.

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