\[\large{ \begin{cases} a^3 = 3(b^2+c^2)-25 \\ b^3=3(c^2+a^2)-25 \\ c^3=3(a^2+b^2)-25 \end{cases}} \]
Let \(a,b\) and \(c\) be distinct real numbers satisfying the system of equations above. Find the product \(abc\).
Note by
Ayush G Rai
2 years, 5 months ago
No vote yet
1 vote
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestSuppose 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 distinct reals, 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 \( (a,b,c) = ( -1, -1 - \sqrt{3}, -1 + \sqrt{3} ) \).
Log in to reply
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 \(\color{white} { f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2). \text{What are the roots?}} \)
Log in to reply
@Calvin Lin Please help me out with the problem!!!!
Log in to reply
Which problem? This one? If so, see the hint.
Log in to reply
How to toggle latex??
Log in to reply
On browser, click on your profile icon
But otherwise, (since the above is already a spoiler space), the second hint says
Log in to reply
Sir it hence means that ABC = 2
Log 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 \[-27+3abc=-21\to abc=2\]
Log in to reply
Awesome solution...(+1) one of the best.
Log in to reply
The answer is \(abc=2\)
Log in to reply
can you post your solution?
Log in to reply
This is from NMTC 2015 lvl 2
Log in to reply
you are absolutely correct.
Log in to reply
What is the sum of 1!+2!+3!+4!+.....+n!
Log in to reply
Actually, there is no neat closed form expression for \(\sum_{n=1}^{k}\,n!\).
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.
Log in to reply
why do you want that?
Log in to reply