How do you use the Binomial Theorem to find the value of #999^3#?

1 Answer
May 28, 2016

Answer:

Rewrite #999^3# as #(10^3-1)^3# and apply the binomial theorem to find that

#999^3=997002999#

Explanation:

The binomial theorem states that #(a+b)^n = sum_(k=0)^n((n),(k))a^(n-k)b^k#

where #((n),(k)) = (n!)/(k!(n-k)!)#

For this problem, we will only need to calculate for #n=3#, and we will find that #((3),(0))=((3),(3))=1# and #((3),(1))=((3),(2))=3#
(Try verifying this)

Noting that it is much easier to calculate powers of #10# and #1# compared to #999#, we can rewrite #999# as #1000-1=10^3-1# and apply the binomial theorem:

#999^3 = (10^3-1)^3#

#=((3),(0))(10^3)^3+((3),(1))(10^3)^2(-1)+((3),(2))(10^3)(-1)^2+((3),(3))(-1)^3#

#=10^9-3*10^6+3*10^3-1#

#=1000000000-3000000+3000-1#

#=997002999#