Question
Compute $(98)^5.$

Answer

Now,we express $98$ as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
Write $98 = 100 – 2$
Thus,$(98)^5 = (100-2)^5$
$=\ ^5C_0 (100)^5 –\ ^5C_1 (100)^4.2 +\ ^5C_2 (100)^32^2 -\ ^5C_3(100)^2 (2)^3 +\ ^5C_4 (100) (2)^4 –\ ^5C_5 (2)^5$
$= 10000000000 – 5 \times 100000000 \times 2 + 10 \times 1000000 \times 4 – 10 \times10000 \times 8 + 5 \times 100 \times 16 – 32$
$= 10040008000 – 1000800032 = 9039207968.$

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