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

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

Note by Ayush Rai
1 year, 3 months ago

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Suppose 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 $$(a,b,c) = ( -1, -1 - \sqrt{3}, -1 + \sqrt{3} )$$. Staff · 1 year, 2 months ago

good simple approach! · 1 year, 2 months ago

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 $$\color{white} { f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2). \text{What are the roots?}}$$ Staff · 1 year, 3 months ago

@Calvin Lin Please help me out with the problem!!!! · 1 year, 3 months ago

Which problem? This one? If so, see the hint. Staff · 1 year, 3 months ago

How to toggle latex?? · 1 year, 3 months ago

On browser, click on your profile icon

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

For a given solution set, consider $$f(x) = x^3 + 3x^2 + 25 - 3(a^2+b^2+c^2).$$ What are the roots?

Staff · 1 year, 3 months ago

Sir it hence means that ABC = 2 · 1 year, 3 months ago

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 $-27+3abc=-21\to abc=2$ · 1 year, 2 months ago

Awesome solution...(+1) one of the best. · 1 year, 2 months ago

The answer is $$abc=2$$ · 1 year, 3 months ago

can you post your solution? · 1 year, 3 months ago

This is from NMTC 2015 lvl 2 · 1 year, 3 months ago

you are absolutely correct. · 1 year, 3 months ago

What is the sum of 1!+2!+3!+4!+.....+n! · 1 year, 3 months ago

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. · 1 year, 2 months ago