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- #1

- Feb 14, 2012

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Evaluate $(a+c)(b+c)(a-d)(b-d)$.

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- #1

- Feb 14, 2012

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Evaluate $(a+c)(b+c)(a-d)(b-d)$.

- Mar 31, 2013

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so $(a+c)(b+c) = (-c - a)(-c-b) = f(-c) = c^2 - 2000c + 1$

as c is root of $x^2-2008x + 1 = 0$ so $c^2 - 2000c +1 = 0$

so $(a+c)(b+c) = 0$ and multiplying by $(a-d)(b-d)$ we get $(a+c)(b+c)(a-d)(b-d) = 0$

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- Feb 14, 2012

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Hi Kali...

Your answer is not correct...I am sorry.

- Mar 31, 2013

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Thanks anemone., I mixed up 2000 and 2008. I have solved it incorrectly

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Evaluate $(a+c)(b+c)(a-d)(b-d)$.

\(\displaystyle (a+c)(b-d)=ab-ad+bc-cd\)

By Viete, we know:

\(\displaystyle ab=cd=1\)

Hence:

\(\displaystyle (a+c)(b-d)=bc-ad\)

Next, let's look at the product:

\(\displaystyle (b+c)(a-d)=ab-bd+ac-cd=ac-bd\)

And so we find:

\(\displaystyle (a+c)(b+c)(a-d)(b-d)=(bc-ad)(ac-bd)=abc^2-b^2cd-a^2cd+abd^2=c^2+d^2-(a^2+b^2)\)

Now, also by Viete, we have:

\(\displaystyle a+b=-2000\)

Hence:

\(\displaystyle a^2+2ab+b^2=2000^2\implies a^2+b^2=2000^2-2\)

Likewise:

\(\displaystyle c+d=2008\)

\(\displaystyle c^2+2cd+d^2=2008^2\implies c^2+d^2=2008^2-2\)

And so:

\(\displaystyle (a+c)(b+c)(a-d)(b-d)=(2008^2-2)-(2000^2-2)=2008^2-2000^2=8(4008)=32064\)

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- Feb 14, 2012

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$\begin{align*}(a+c)(b+c)(a-d)(b-d)&=[ab+(a+b)c+c^2][ab-(a+b)d+d^2]\\&=(1-2000c+c^2)(1+2000d+d^2)\\&=(c^2-2008c+1+8c)(d^2-2008d+1+4008d)\\&=8c(4008d)\\&=32064\end{align*}$