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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink]
Bunuel wrote:
Which of the following is equivalent to \(\frac{{{a}^{100}}-{{b}^{100}}}{{{a}^{50}}-{{b}^{50}}}\) for all values of a and b for which the expression is defined?

(A) a^2+b^2
(B) a^2-b^2
(C) a^50+b^50
(D) a^50-b^50
(E) (ab)^2

Kudos for a correct solution.


\(\frac{{{a}^{100}}-{{b}^{100}}}{{{a}^{50}}-{{b}^{50}}}\)

= \(\frac{(a^{50})^2 - (b^{50})^2}{a^{50}-b^{50}}\)

= \(\frac{(a^{50}+b^{50})(a^{50}-b^{50})}{a^{50}-b^{50}}\)

= \(a^{50}+b^{50}\)

Answer:- C
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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink]
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Bunuel wrote:
Which of the following is equivalent to \(\frac{{{a}^{100}}-{{b}^{100}}}{{{a}^{50}}-{{b}^{50}}}\) for all values of a and b for which the expression is defined?

(A) a^2+b^2
(B) a^2-b^2
(C) a^50+b^50
(D) a^50-b^50
(E) (ab)^2

Kudos for a correct solution.


To solve this we have to remember the basic formula : \(a^2 - b^2 = (a+b)(a-b)\)

Here the numerator \(a^{100}-b^{100}\) Can be changed to \((a^{50})^2 -(b^{50})^2\)

Clearly this can now be re written as \(({a}^{50}+{b}^{50})({a}^{50}-{{b}^{50})\)

Simplify this with the denominator whats left is \({a}^{50}+{b}^{50}\) and thus the Answer is C
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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink]
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[(a^50 - b^50)(a^50 + b^50)] / (a^50 - b^50)

Answer is (a^50 + b^50) = C
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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink]
Bunuel wrote:
Which of the following is equivalent to \(\frac{{{a}^{100}}-{{b}^{100}}}{{{a}^{50}}-{{b}^{50}}}\) for all values of a and b for which the expression is defined?


(A) \(a^2+b^2\)

(B) \(a^2-b^2\)

(C) \(a^{50}+b^{50}\)

(D) \(a^{50}-b^{50}\)

(E) \((ab)^2\)


Lets \(a^{50} = x\) and \(b^{50} = y\)

Now \(\frac{{{a}^{100}}-{{b}^{100}}}{{{a}^{50}}-{{b}^{50}}}\) = \(\frac{{{x}^{2}}-{{y}^{2}}}{x-y}\) = \(\frac{(x+y) (x-y) }{ (x-y)}\) = (x+Y)

= \(a^{50} + b^{50}\)

Ans C
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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink]
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