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Exponent Challenge

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anemone

MHB POTW Director
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Feb 14, 2012
3,681
Prove that \(\displaystyle \left( 6+845^{\frac{1}{3}}+325^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 6+847^{\frac{1}{3}}+539^{\frac{1}{3}} \right)^{\frac{1}{3}}=\left( 4+245^{\frac{1}{3}}+175^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 8+1859^{\frac{1}{3}}+1573^{\frac{1}{3}} \right)^{\frac{1}{3}}\)
 

Opalg

MHB Oldtimer
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Feb 7, 2012
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Let $\alpha = \sqrt[3]5,\ \beta = \sqrt[3]7,\ \gamma = \sqrt[3]11,\ \delta = \sqrt[3]13,$ and let $\epsilon = \sqrt[3]{1/3}.$ Then $845 = 5\times13^2$ and $325 = 5^2\times 13$. So $$( 6+845^{1/3}+325^{1/3})^{1/3} = ( 6+ \alpha\delta^2 + \alpha^2\delta)^{1/3} = \epsilon( 13+ 3\alpha\delta^2 + 3\alpha^2\delta + 5)^{1/3} = \epsilon( (\delta+ \alpha)^3)^{1/3} = \epsilon(\delta+ \alpha).$$ In exactly the same way, $( 6+847^{1/3}+539^{1/3})^{1/3} = \epsilon(\gamma+ \beta)$, $( 4+245^{1/3}+175^{1/3})^{1/3} = \epsilon(\beta+ \alpha)$ and $ (8+1859^{1/3}+1573^{1/3} )^{1/3} = \epsilon(\delta+ \gamma)$. Thus the problem reduces to showing that $\epsilon(\delta+ \alpha) + \epsilon(\gamma+ \beta) = \epsilon(\beta+ \alpha) + \epsilon(\delta+ \gamma)$, which is obviously true.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,681
Prove that \(\displaystyle \left( 6+845^{\frac{1}{3}}+325^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 6+847^{\frac{1}{3}}+539^{\frac{1}{3}} \right)^{\frac{1}{3}}=\left( 4+245^{\frac{1}{3}}+175^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 8+1859^{\frac{1}{3}}+1573^{\frac{1}{3}} \right)^{\frac{1}{3}}\)
Thank you Opalg for participating and the nice approach!:)

My solution:

Notice that
\(\displaystyle \left( 6+845^{\frac{1}{3}}+325^{\frac{1}{3}} \right)^{\frac{1}{3}}=( 6+5^{\frac{1}{3}}13^{\frac{2}{3}} +5^{\frac{2}{3}}13^{\frac{1}{3}} )^{\frac{1}{3}}\) can be rewritten as the sum of two cube roots, i.e.
\(\displaystyle
( 6+5^{\frac{1}{3}}13^{\frac{2}{3}} +5^{\frac{2}{3}}13^{\frac{1}{3}} )^{\frac{1}{3}}=a^{\frac{1}{3}}+b^{\frac{1}{3}}\), $a, b \in Z$

Raise both sides of the equation to the third power and by comparing the bases of the two exponents, we get:

\(\displaystyle
( 6+5^{\frac{1}{3}}13^{\frac{2}{3}} +5^{\frac{2}{3}}13^{\frac{1}{3}} )^{\frac{1}{3}}=(a^{\frac{1}{3}}+b^{\frac{1}{3}})^{\frac{1}{3}}\)

\(\displaystyle
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a+b+3(ab)^{ \frac{1}{3}}(a^{\frac{1}{3}}+b^{\frac{1}{3}})\)

\(\displaystyle
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a+b+(3b)^{ \frac{1}{3}}(3a)^{ \frac{2}{3}}+(3a)^{ \frac{1}{3}}(3b)^{ \frac{2}{3}}\)

We see that \(\displaystyle a=\dfrac{13}{3}\), \(\displaystyle b=\dfrac{5}{3}\) and hence

\(\displaystyle \left( 6+845^{\frac{1}{3}}+325^{\frac{1}{3}} \right)^{\frac{1}{3}}=( 6+5^{\frac{1}{3}}13^{\frac{2}{3}} +5^{\frac{2}{3}}13^{\frac{1}{3}} )^{\frac{1}{3}}=\left(\dfrac{13}{3} \right)^{ \frac{1}{3}}+ \left(\dfrac{5}{3} \right)^{ \frac{1}{3}}\)

Similarly,

\(\displaystyle \left( 6+847^{\frac{1}{3}}+539^{\frac{1}{3}} \right)^{\frac{1}{3}}=( 6+7^{\frac{1}{3}}11^{\frac{2}{3}} +11^{\frac{2}{3}}7^{\frac{1}{3}} )^{\frac{1}{3}}=\left(\dfrac{11}{3} \right)^{ \frac{1}{3}}+ \left(\dfrac{7}{3} \right)^{ \frac{1}{3}}\)

\(\displaystyle \left( 4+245^{\frac{1}{3}}+175^{\frac{1}{3}} \right)^{\frac{1}{3}}=( 4+5^{\frac{1}{3}}7^{\frac{2}{3}} +5^{\frac{2}{3}}7^{\frac{1}{3}} )^{\frac{1}{3}}=\left(\dfrac{7}{3} \right)^{ \frac{1}{3}}+ \left(\dfrac{5}{3} \right)^{ \frac{1}{3}}\)

\(\displaystyle \left( 8+1859^{\frac{1}{3}}+1573^{\frac{1}{3}} \right)^{\frac{1}{3}}=( 4+5^{\frac{1}{3}}7^{\frac{2}{3}} +5^{\frac{2}{3}}7^{\frac{1}{3}} )^{\frac{1}{3}}=\left(\dfrac{13}{3} \right)^{ \frac{1}{3}}+ \left(\dfrac{11}{3} \right)^{ \frac{1}{3}}\)

Therefore the required result follows.