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If in the expansion of $(1 + x)^m \cdot (1 – x)^n $, the coefficients of $ x $ and $ x^2 $are 3 and -6 respectively, then m is ?

I solved it in the following way :

Expanding we get, the coefficient of $ x $ as $ (m-n) = 3 \cdots (1)$ and coefficient of $ x^2 $ as $ \frac{n(n-1)}{2} + \frac{m(m-1)}{2} +- m \cdot n = -6 \cdots (2)$ after substituting and some algebraic treatment I got m = 12 and n = 9.

Now this is correct but I am interested if there exists any short procedure such that the entire problem could be done under a mint.

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    The only thing that jumps out at me is the appearance of a convenient square trinomial: clearing the fractions and expanding your equation (2), we get $n^2 - n + m^2 - m - 2 m n = -12$. The second-degree terms combine nicely: $n^2+m^2-2mn = (n-m)^2$, and your equation (1) tells us we can replace this value with $3^2$. This leaves you with the system $m-n=3$ and $m+n=21$, which is very easy to solve.2010-10-26
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    I didn't understand what you meant , precisely `clearing the fractions and expanding your equation (2), we get n2−n+m2−m−2mn=−12.` ?!2010-10-26
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    If you write small enough, you could perhaps fit the solution under a mint. But I think you mean "under a minute". ;-)2010-10-26
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    @Tretwick I should probably have written "denominators", not "fractions". (Too late to edit my comment.) If you multiply each term of your equation (2) by 2, you get $n(n-1) + m(m-1) - 2mn = -12$; on the left-hand side, this has "cleared the denominators", which tends to make an equation like this easier to handle. (Fractions can be a little intimidating ... but now we don't have any!) By "expanding", I simply mean to multiply-out the terms $n(n-1)$ and $m(m-1)$, which gives the equation $n^2-n+m^2-m-2mn=-12$.2010-10-26
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    Instead of binomial formula you can also derivate and plugin $x=0$ ;)2011-12-30

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