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Question 269183: (2r^(-1)+ s^(-1))^5 Expand the sum of the two terms of a Binomial. Using the Binomial Theorem.
Answer by persian52(161)   (Show Source):
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(2r^(-1)+s^(-1))^(5)
Remove the negative exponent in the numerator by rewriting 2r^(-1) as (2)/(r). A negative exponent follows the rule: a^(-n)=(1)/(a^(n)).
((2)/(r)+s^(-1))^(5)
Remove the negative exponent in the numerator by rewriting s^(-1) as (1)/(s). A negative exponent follows the rule: a^(-n)=(1)/(a^(n)).
((2)/(r)+(1)/(s))^(5)
To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is rs. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
((2)/(r)*(s)/(s)+(1)/(s)*(r)/(r))^(5)
Complete the multiplication to produce a denominator of rs in each expression.
((2s)/(rs)+(r)/(rs))^(5)
Combine the numerators of all expressions that have common denominators.
((2s+r)/(rs))^(5)
Expand the exponent of 5 to the inside factor (2s+r).
((2s+r)^(5))/((rs)^(5))
Expand the exponent 5 to rs.
Answer: ((2s+r)^(5))/(r^(5)s^(5))
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