MHB Solve Exponent Challenge: Prove Equality

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The discussion focuses on proving the equality of two complex expressions involving cube roots. The key simplification involves substituting specific values for the cube roots of integers, leading to a reduction of the original equation to a simpler form. By demonstrating that both sides of the equation can be expressed in terms of the same variables, the proof is established. The final step confirms that the simplified expressions on both sides are equal, validating the original equation. The collaborative effort highlights the effectiveness of algebraic manipulation in solving exponent challenges.
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Prove that $$\left( 6+845^{\frac{1}{3}}+325^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 6+847^{\frac{1}{3}}+539^{\frac{1}{3}} \right)^{\frac{1}{3}}=\left( 4+245^{\frac{1}{3}}+175^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 8+1859^{\frac{1}{3}}+1573^{\frac{1}{3}} \right)^{\frac{1}{3}}$$
 
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[sp]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.[/sp]
 
anemone said:
Prove that $$\left( 6+845^{\frac{1}{3}}+325^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 6+847^{\frac{1}{3}}+539^{\frac{1}{3}} \right)^{\frac{1}{3}}=\left( 4+245^{\frac{1}{3}}+175^{\frac{1}{3}} \right)^{\frac{1}{3}}+\left( 8+1859^{\frac{1}{3}}+1573^{\frac{1}{3}} \right)^{\frac{1}{3}}$$

Thank you Opalg for participating and the nice approach!:)

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

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

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

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

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

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

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

Similarly,

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

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

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

Therefore the required result follows.
 
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