How do you use the binomial theorem to expand and simplify the expression #(3a-b)^5#?

1 Answer

Answer:

#(3a-b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5#

Explanation:

Binomial expansion is done using:

#(a+b)^n=(C_(n,0))a^nb^0+(C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n#

And so we have here:

#(3a-b)^5=(C_(5,0))a^5b^0+(C_(5,1))a^4b^1+(C_(5,2))a^3b^2+(C_(5,3))a^2b^3+(C_(5,4))a^1b^4+(C_(5,5))a^0b^5#

#(3a-b)^5=(1)a^5b^0+(5)a^4b^1+(10)a^3b^2+(10)a^2b^3+(5)a^1b^4+(1)a^0b^5#

#(3a-b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5#