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03 Mar 2019, 15:46
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C
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Difficulty:
15%
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Question Stats:
83% (01:50) correct
17%
(01:31)
wrong
based on 72
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GMATH practice exercise (Quant Class 3)
If \(\left( {a,b} \right) = \left( {9,8} \right)\) , the expression \({{2\left( {{a^3} - {b^3}} \right)} \over {a\left( {a + b} \right) + {b^2}}}\) is:
(A) negative
(B) zero
(C) between 0 and 3
(D) between 3 and 6
(E) greater than 6
If \(\left( {a,b} \right) = \left( {9,8} \right)\) , the expression \({{2\left( {{a^3} - {b^3}} \right)} \over {a\left( {a + b} \right) + {b^2}}}\) is:
(A) negative
(B) zero
(C) between 0 and 3
(D) between 3 and 6
(E) greater than 6
03 Mar 2019, 16:14
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Keep in mind that:
\({a^3} - {b^3} = (a - b) ({a^2} + ab + {b^2})\)
Hence:
\(2 \frac{(a^3 - b^3)}{a (a + b) + b^2} = 2 (a - b) \frac{(a^2 + ab + b^2)}{a^2 + ab + b^2} = 2 (9 - 8) = 2\)
Thus, the correct answer is C.
\({a^3} - {b^3} = (a - b) ({a^2} + ab + {b^2})\)
Hence:
\(2 \frac{(a^3 - b^3)}{a (a + b) + b^2} = 2 (a - b) \frac{(a^2 + ab + b^2)}{a^2 + ab + b^2} = 2 (9 - 8) = 2\)
Thus, the correct answer is C.
03 Mar 2019, 16:39
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fskilnik
\(?\,\, = \,\,\frac{{2\left( {{a^3} - {b^3}} \right)}}{{a\left( {a + b} \right) + {b^2}}}\,\,\,\,\,{\text{for}}\,\,\,\,\left( {a,b} \right) = \left( {9,8} \right)\)
\(\left. \matrix{\\
2\left( {{a^3} - {b^3}} \right) = 2\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\,\,\, \hfill \cr \\
a\left( {a + b} \right) + {b^2} = {a^2} + ab + {b^2} \hfill \cr} \right\}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,{{2\left( {{a^3} - {b^3}} \right)} \over {{a^2} + ab + {b^2}}} = 2\left( {a - b} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {a,b} \right) = \left( {9,8} \right)} \,\,\,\,\,? = 2\)
\(\left( * \right)\,\,{\rm{when}}\,\,\,{a^2} + ab + {b^2} \ne 0\)
The correct answer is (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
