1) Determine all pairs of real numbers \((x,y)\) such that

\(x^6 = y^4 + 18\)

\(y^6 = x^4 + 18\)

2) Let \(a,b,c\) be integers such that \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=3\). Prove that \(abc\) is a perfect cube.

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## Comments

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TopNewestFirst of all replace \(({x}^2, {y}^2)\) by a and b. Then it is easy to see that if \(a \geq b\) then \( {a}^3 \geq {b}^3\) but that would also mean that \( {a}^2 \leq {b}^2\) comparing the rhs of the two equations . But we have \(a \geq b\) so we also have \({a}^2 \geq {b}^2\) . Hence a = b now we write the equation as \( {a}^3 - {a}^2 - 18 = 0 \) it is easy to see that a = 3 satisfies the equation and so we get \( x = y = \pm \sqrt {3}\) we then get another quadratic factor whose roots are obtained and then we get two more values of x as ;- \( \sqrt {-1 \pm \sqrt{-5}}\) so we have only two real solutions x = y = \( \pm \sqrt{3}\)

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Nice way @akash deep

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For question no.2

By AM-GM we see that:-

\[\Large{\frac { a }{ b } +\frac { b }{ c } +\frac { c }{ a } \ge 3\sqrt [ 3 ]{ \frac { a }{ b } \times \frac { b }{ c } \times \frac { c }{ a } } =3}\]

So we see the minimum value of expression and its value are same therefore \(a=1,b=1,c=1\) for the equality.

We see \(abc=1\) which is a perfect cube!

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I too did it the same way but the question is not restricted to natural numbers it is for integers which can be negative also. So we can't use AM -GM .

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Yes that's a problem

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Comment deleted Nov 11, 2015

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Are you going celebrated Diwali today with crackers

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@Lakshya Sinha you??

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@Harsh Shrivastava , please check it. @Dev Sharma

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@Pi Han Goh @Nihar Mahajan @Surya Prakash Adarsh Kumar

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@Swapnil Das @Kushagra Sahni

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I have posted a question related to the first one with the name; "inspired by dev sharma's note" u can check that out.

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Hey can u plz post the solution for the second question for the cases in which a,b,c may be negative. I have done it upto natural numbers. Please post the solution.

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