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23 Dec 2019, 23:20
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C
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If \(x + \frac{2}{x} = 3\) what is the value of \(x^3 + \frac{8}{x^3} =\) ?
A. 1
B. 8
C. 9
D. 16
E. 18
A. 1
B. 8
C. 9
D. 16
E. 18
24 Dec 2019, 00:54
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Bunuel
If you do not know the algebraic formula, work on the equation and you will get your answer ..
\(x + \frac{2}{x} = 3.........x^2+2=3x.........x^2-3x+2=0....(x-1)(x-2)=0\)
so x=1 or x=2
Substitute and there is your answer..
\(x^3 + \frac{8}{x^3} =\)\(1^3 + \frac{8}{1^3} =1+8=9\)
C
General Discussion
23 Dec 2019, 23:39
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we are given \(x + \frac{2}{x}\) and we need the value of \(x^3 + \frac{8}{x^3}\)
we apply the the identity \((a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2\)
\((x +\frac{2}{x})^3 = x^3 + \frac{8}{x^3} + 3.x^2.\frac{2}{x} +3.x.\frac{4}{x^2} \)
we know that \(x + \frac{2}{x} = 3\)
\(27 = x^3 + \frac{8}{x^3} + 6x + \frac{12}{x}\)
\(27 = x^3 + \frac{8}{x^3} + 6(x + \frac{2}{x})\)
\(27 = x^3 + \frac{8}{x^3} + 6(3)\)
\(9 = x^3 + \frac{8}{x^3}\)
IMO,C
we apply the the identity \((a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2\)
\((x +\frac{2}{x})^3 = x^3 + \frac{8}{x^3} + 3.x^2.\frac{2}{x} +3.x.\frac{4}{x^2} \)
we know that \(x + \frac{2}{x} = 3\)
\(27 = x^3 + \frac{8}{x^3} + 6x + \frac{12}{x}\)
\(27 = x^3 + \frac{8}{x^3} + 6(x + \frac{2}{x})\)
\(27 = x^3 + \frac{8}{x^3} + 6(3)\)
\(9 = x^3 + \frac{8}{x^3}\)
IMO,C
Bunuel
