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If (a,b)=(9,8), the expression 2(a^3-b^3)/(a(a+b)+b^2) is:

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by fskilnik@GMATH » Sun Mar 03, 2019 2:35 pm

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A

B

C

D

E

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GMATH practice exercise (Quant Class 3)

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Answer: [spoiler]_____(C)__[/spoiler]
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If (a,b)=(9,8), the expression 2(a^3-b^3)/(a(a+b)+b^2) is:

by fskilnik@GMATH » Sun Mar 03, 2019 3:30 pm
fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 3)

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$$?\,\, = \,\,{{2\left( {{a^3} - {b^3}} \right)} \over {a\left( {a + b} \right) + {b^2}}}\,\,\,\,\,{\rm{for}}\,\,\,\,\left( {a,b} \right) = \left( {9,8} \right)$$

$$\left. \matrix{
2\left( {{a^3} - {b^3}} \right) = 2\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\,\,\, \hfill \cr
a\left( {a + b} \right) + {b^2} = {a^2} + ab + {b^2} \hfill \cr} \right\}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,{{2\left( {{a^3} - {b^3}} \right)} \over {{a^2} + ab + {b^2}}} = 2\left( {a - b} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {a,b} \right) = \left( {9,8} \right)} \,\,\,\,\,? = 2$$
$$\left( * \right)\,\,{\rm{when}}\,\,\,{a^2} + ab + {b^2} \ne 0$$


The correct answer is (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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