Question 394655
Pascal's triangle is very useful for finding the 
coefficients of any expression of the form
{{{(a + b)^k}}} , where {{{k = 1,2,3, . . .}}} etc.
Each entry is the sum of the entries on the 
previous line on the left and right.
the triangle looks like:

1 - - - - -  1
1 - -  2-  - 1
1 - 3 - -3 -1
1 4 - 6 -   4 1
It's hard to show on the computer, but
the next line is the one you need:
1 - 5 - 10 - 10 - 5 - 1
These are the coefficients of the 
expansion of {{{(a + b)^5}}}
It tells you there are 6 terms in
the expansion.
The first term is {{{a^5}}}, and the coefficient is {{{1}}}
The last term is {{{b^5}}}, and the coefficient is also {{{1}}}
The whole expansion looks like
{{{a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5a*b^4 + b^5}}}
You may wonder what this does for you.
Well, I you just say
{{{a = x}}} and
{{{b = -7}}}, your problem is solved
{{{x^5 + 5x^4*(-7) + 10x^3*(-7)^2 + 10x^2*(-7)^3 + 5x*(-7)^4 + b^5}}} 
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Here's a list of the nasty powers of {{{-7}}}:
{{{(-7)^2 = 49}}}
{{{(-7)^3 = -343}}}
{{{(-7)^4 = 2401}}}
{{{(-7)^5 = -16807}}}
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{{{x^5 - 35x^4 + 490x^3 - 3430x^2 + 12005x - 16807}}}  answer
Hope this helps