How do you use the Binomial Theorem to expand $(x-3)^5$ ?

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Answer 1 · 2016-06-30T18:26:50

$(x-3)^5=x^5-15x^4+90x^3-270x^2+405x-243.$

The Binomial Theorem Expansion $: (x+y)^n=x^n+""nC_1x^(n-1)y+""nC_2x^(n-2)y^2+""nC_3x^(n-3)y^3+.........+""nC_rx^(n-r)y^r+... ...+y^n.$

Letting $y=-3, n=5$ , we get,
$(x-3)^5=x^5+""5C_1x^4(-3)^1+""5C_2x^3(-3)^2+""5C_3x^2(-3)^3+""5C_4x^1(-3)^4+(-3)^5.$

Here, $""5C_1=""5C_(5-1)=""5C_4=5, ""5C_2=""5C_3=(5*4)/(1*2)=10$

So, $(x-3)^5=x^5+5x^4(-3)+10x^3(9)+10x^2(-27)+5x(81)+(-243).$
$=x^5-15x^4+90x^3-270x^2+405x-243.$

How do you use the Binomial Theorem to expand (x-3)^5(x-3)^5?