\[\large{ \begin{cases} a^3 = 3(b^2+c^2)-25 \\ b^3=3(c^2+a^2)-25 \\ c^3=3(a^2+b^2)-25 \end{cases}} \]

Let \(a,b\) and \(c\) be distinct real numbers satisfying the system of equations above. Find the product \(abc\).

\[\large{ \begin{cases} a^3 = 3(b^2+c^2)-25 \\ b^3=3(c^2+a^2)-25 \\ c^3=3(a^2+b^2)-25 \end{cases}} \]

Let \(a,b\) and \(c\) be distinct real numbers satisfying the system of equations above. Find the product \(abc\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestSuppose that we have a solution of distinct reals \( (a,b,c) \). Let \( S = a^2 + b^2 + c^2 \).

Then, we have \( a^3 + 3a^2 + 25 = 3 (a^2 + b^2 + c^2 ) = 3S \), and similar equations for \(b\) and \(c\).

Since these are

distinctreals, we know that the roots to the cubic equation \( X^3 + 3X^2 + 25 - 3S = 0 \) must be \(a, b, c \).Hence, by vieta's formula, this tells us \( a+b+c = -3, ab+bc+ca = 0 \).

Thus, \( S = (a^2+b^2+c^2) = (a+b+c)^2 - 2(ab+bc+ca) = 9 \). Finally, using the constant coefficient, \( abc = - ( 25 - 3S ) = - (25 - 27) = 2 \).

Note: We can now solve \( x^3 + 3x^2 -2 = 0 \) to obtain that \( (a,b,c) = ( -1, -1 - \sqrt{3}, -1 + \sqrt{3} ) \). – Calvin Lin Staff · 10 months ago

Log in to reply

– Ayush Rai · 10 months ago

good simple approach!Log in to reply

Hint:What can we say about \( a^3 + 3a^2 + 25 \)?The next hint is in white text, so you will need to Toggle Latex.

Hint:For a given solution set, consider \(\color{white} { f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2). \text{What are the roots?}} \) – Calvin Lin Staff · 10 months, 1 week agoLog in to reply

@Calvin Lin Please help me out with the problem!!!! – Ankit Kumar Jain · 10 months, 1 week ago

Log in to reply

– Calvin Lin Staff · 10 months, 1 week ago

Which problem? This one? If so, see the hint.Log in to reply

– Ankit Kumar Jain · 10 months, 1 week ago

How to toggle latex??Log in to reply

But otherwise, (since the above is already a spoiler space), the second hint says

– Calvin Lin Staff · 10 months, 1 week agoLog in to reply

– Ankit Kumar Jain · 10 months, 1 week ago

Sir it hence means that ABC = 2Log in to reply

first subtract the equation simultaneously to get \[\begin{cases} a^3-b^3=-3(a^2-b^2)\\ b^3-c^3=-3(b^2-c^2)\\ c^3-a^3=-3(c^2-a^2) \end{cases}\] since all variable are distinct we can divide both side by (a-b),(b-c),(c-a) respectively. we get \[\begin{cases} a^2+ab+b^2+3(a+b)=0\\ b^2+bc+c^2+3(b+c)=0\\ c^2+ca+c^2+3(c+a)=0\\ \end{cases}\] subtract any two equation to get \[a+b+c=-3\] add all the equations to get \[2(a^2+b^2+c^2)+ab+bc+ca+6(a+b+c)=0\] using the fact\(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-2(ab+bc+ca)\) \[ab+bc+ca=0\] adding all the original equations \[a^3+b^3+c^3=6(a^2+b^2+c^2)-75=-21\] using the identity \[a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=-27+3abc\] so we have \[-27+3abc=-21\to abc=2\] – Aareyan Manzoor · 10 months ago

Log in to reply

– Ayush Rai · 10 months ago

Awesome solution...(+1) one of the best.Log in to reply

The answer is \(abc=2\) – Janardhanan Sivaramakrishnan · 10 months, 1 week ago

Log in to reply

– Ayush Rai · 10 months, 1 week ago

can you post your solution?Log in to reply

This is from NMTC 2015 lvl 2 – Ankit Kumar Jain · 10 months, 1 week ago

Log in to reply

– Ayush Rai · 10 months, 1 week ago

you are absolutely correct.Log in to reply

What is the sum of 1!+2!+3!+4!+.....+n! – Bunneng Nath · 10 months, 1 week ago

Log in to reply

There are some formulae but you would need to know about exponential integrals and several other advance concepts for understanding them.

However, there are some series involving factorials like \(\sum_{n=1}^{k}\,n \cdot n!\) which can be described by simple expressions. – Aditya Sky · 10 months ago

Log in to reply

– Ayush Rai · 10 months, 1 week ago

why do you want that?Log in to reply