Binomial Expansion

(a) Expand (1+2x)^6 (1-3x)^6 in ascending powers of x as far as the term containing x^3.
(1+2x)^6=1+12x+60x^2+160x^3+...
(1-3x)^6=1-18x+135x^2-540x^3+...
(1+2x)^6(1-3x)^6
= (1+12x+60x^2+160x^3+...)(1-18x+135x^2-540x^3+...)
= 1+(12-18)x+(135+60-18*12)x^2+(160-540+135*12-18*60)x^3+...
= 1-6x-21x^2+160x^3+...
(b) Hence, evaluate 0.98^6 and correct your answer to the 4 decimal places
0.98^6
=(1-0.02)^6
=(1-2/100)^6
Let x = -1/100,
Hence, we get 0.98^6 = (1+2x)^6
= 1+12x+60x^2+160x^3+240x^4+192x^5+64x^6
= 1+12(-1/100)+60(-1/100)^2+160(-1/100)^3+240(-1/100)^4+...
= 1-12/100+60/10000-160/1000000+...
= 1-0.12+0.006-0.00016+...
= 0.88584+...
=0.8858(corr to 4 sig.fig.)