MHB Evaluate the value of the product

[anemone]

The three pairs of roots $(a,\,b)$ that satisfy $a^3-3ab^2=2005$ and $b^3-3b^2a=2004$ are $(a_1,\,b_1),\,(a_2,\,b_2),\,(a_3,\,b_3)$.

Evaluate $\left(\dfrac{b_3-a_3}{b_3}\right)\left(\dfrac{b_2-a_2}{b_2}\right)\left(\dfrac{b_1-a_1}{b_1}\right)$.

 

[Albert1]

Spoiler

$a^3-3ab^2=2005---(1)$
$b^3-3b^2a=2004---(2)$
(2)-(1) and simplify we get :$(b-a)^3=-1,\,\ or\\, (b-a)=-1--(3)$
put (3) to (2) we get :$2b^3+3b^2+2004=0---(4)$
$b_3-a_3=b_2-a_2=b_1-a_1=b-a=-1$
and $b_1b_2b_3=-1002$
$\therefore $ the answer =$\dfrac {1}{1002}$

 

[lfdahl]

Spoiler

I have a question to Albert, because I do not quite understand his deduction:
The equations
\[a^3-3ab^2 = 2005\: \: \: \: \: (1). \\\\ b^3-3b^2a = 2004\: \: \: \: \: (2).\]
lead to:
\[b^3-a^3 = -1 \: \: \:\: (3).\]
and not to:
\[(b-a)^3 = -1\]
- because there is not the term $3a^2b$ in either (1) or (2)??
Thanks for clearing the matter

 

[Albert1]

I think (1) should be :$a^3-3a^2b=2005$

 

[anemone]

Ops...I previously received a PM that notified me of the possible typo that I could have made, but I thought he mentioned of the constant value, apparently Albert was right, the first equation has a typo in it, and his intuition was right as well.

Sorry for both the late reply and late clarification post. :(

 

[kaliprasad]

Hello Anemone,
I have solved it here

Spoiler

Fun with maths: 2015/014) if $a^3-3a^2b = 2005$ and $b^3-3b^2a = 2004$ then find

 

[MarkFL]

Hello Anemone,
I have solved it here

Spoiler

Fun with maths: 2015/014) if $a^3-3a^2b = 2005$ and $b^3-3b^2a = 2004$ then find

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