×

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
1 year, 4 months ago

Sort by:

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 = $$\pm \sqrt{3}$$ · 1 year, 4 months ago

Nice way @akash deep · 1 year, 4 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! · 1 year, 4 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 . · 1 year, 4 months ago

Yes that's a problem · 1 year, 4 months ago

Comment deleted Nov 11, 2015

Are you going celebrated Diwali today with crackers · 1 year, 4 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 · 1 year, 4 months ago

@Lakshya Sinha you?? · 1 year, 4 months ago

Nope it's waste of time · 1 year, 4 months ago

Come on google + hangout · 1 year, 4 months ago

AGREEEE · 1 year, 4 months ago

Well I have uninstalled Hangouts, give me your email id I will add you on circles in Google+ · 1 year, 4 months ago

sdev3377@gmail.com · 1 year, 4 months ago

I think you added me in your circles so did I · 1 year, 4 months ago

@Harsh Shrivastava , please check it. @Dev Sharma · 1 year, 4 months ago

@Pi Han Goh @Nihar Mahajan @Surya Prakash Adarsh Kumar · 1 year, 4 months ago

@Swapnil Das @Kushagra Sahni · 1 year, 4 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. · 1 year, 4 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. · 1 year, 4 months ago

×