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.

Note by Dev Sharma
2 years, 6 months ago

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First 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 = $$\pm \sqrt{3}$$

- 2 years, 6 months ago

Nice way @akash deep

- 2 years, 6 months ago

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!

- 2 years, 6 months ago

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 .

- 2 years, 6 months ago

Yes that's a problem

- 2 years, 6 months ago

Comment deleted Nov 11, 2015

Are you going celebrated Diwali today with crackers

- 2 years, 6 months ago

NO. Today, I would celebrate it without crackers... I would not fire crackers because it causes pollution and I dont find any good reason to fire them

- 2 years, 6 months ago

@Lakshya Sinha you??

- 2 years, 6 months ago

Nope it's waste of time

- 2 years, 6 months ago

- 2 years, 6 months ago

AGREEEE

- 2 years, 6 months ago

Well I have uninstalled Hangouts, give me your email id I will add you on circles in Google+

- 2 years, 6 months ago

sdev3377@gmail.com

- 2 years, 6 months ago

- 2 years, 6 months ago

@Harsh Shrivastava , please check it. @Dev Sharma

- 2 years, 6 months ago

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

- 2 years, 6 months ago

- 2 years, 6 months ago

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

- 2 years, 6 months ago

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.

- 2 years, 6 months ago