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Which of the following is equivalent to (a^100-b^100)/(a^50-b^50)

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Bunuel
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Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink] New post   23 Sep 2015, 22:46
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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) \((ab)^2\)
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BrentGMATPrepNow
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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink] New post   03 Oct 2018, 10:26
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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.


The numerator has a nice difference of squares, which can be factored
Note: a^100 = (a^50)²

So, (a^100 - b^100)/(a^50 - b^50) = (a^50 + b^50)(a^50 - b^50)/(a^50 - b^50)
= a^50 + b^50

Answer: C
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anudeep133
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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink] New post   24 Sep 2015, 01:26
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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.


Solution:
\(\frac{{{a}^{2*50}}-{{b}^{2*50}}}{{{a}^{50}}-{{b}^{50}}}\) = \(\frac{({{a}^{50}}-{{b}^{50}})({{a}^{50}}+{{b}^{50}})}{{{a}^{50}}-{{b}^{50}}}\) = \({{a}^{50}}+{{b}^{50}}\)

Option C
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Re: Which of the following is equivalent to (a^100-b^100)/(a^50-b^50) [#permalink] New post   25 Sep 2015, 23:49
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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] New post   27 Sep 2015, 00:22
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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] New post   24 Sep 2015, 06:54
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] New post   23 Feb 2021, 09:58
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]
23 Feb 2021, 09:58
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