"Use the Binomial Theorem to expand each binomial."
(x-3)^5
How can I solve this without using the aid of Pascal's Triangle?
(The answer, for reference, is x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243.)

Question

"Use the Binomial Theorem to expand each binomial." 
(x-3)^5
How can I solve this without using the aid of Pascal's Triangle?
(The answer, for reference, is x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243.)

solomonzelman · Answer

http://www.intmath.com/series-binomial-theorem/4-binomial-theorem.php

anonymous · Answer

$$\left( x+a \right)^n=nc0 x^n+nc1x ^{n-1}a+nc2 x ^{n-2}a^2+nc3x ^{n-3}a^2+$$
$$nc4x ^{n-4}a^4+nc5 x ^{n-5}a^5+...+ncrx ^{n-r}a^r+...+ncn a^n$$
nc0=1$$nc1=\frac{ n }{ 1 },nc2=\frac{ n \left( n-1 \right) }{ 2*1 },nc3=\frac{ n \left( n-1 \right)\left( n-2 \right) }{  3*2*1 },...$$
here a=-3,n=5
solve