We have to use the binomial theorem
#(a+b)^n = a^n b^0+na^{n-1}b+{n(n-1)}/{2!}x^2+...+a^0b^n#
with #a=1#, #b=-3x# and #n=8#. Although the binomial expansion has 9 terms in this case, we retain only the first 4 to get the expansion up to #x^3#
#(1-3x)^8 = 1+8times (-3x)+{8times 7}/{1times 2}(-3x)^2+{8times 7 times 6}/{1 times 2 times 3}(-3x)^3+...#
#=1-24x+252x^2-1512x^3+....#
If we use this to calculate #0.97^8# we need to take #x= 0.01# in this formula
#0.97^8 = (1-3times 0.01)^8 ~~ 1-24times 0.01 +252 times 0.01^2 - 1512times 0.01^3 = 0.783688 #
Comparing this with the exact value 0.783743359437696, we see that our approximation is correct to 4 decimal places (this was expected, since the next term in the binomial expansion
#{8 times 7 times 6 times 5}/{1 times 2 times 3 times 4}(-3x^4) = 5670x^4# is smaller than #10^{-4}# for #x=0.01#). The value, rounded off to four decimal places is 0.7837
