If x + 1/16x = 1, then the value of 64 x^3 + 1/64^3 is

a)4

b)52

c)64

d)76

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- Jun 16th 2013, 04:45 AMsscgeekQuadratic equation question
If x + 1/16x = 1, then the value of 64 x^3 + 1/64^3 is

a)4

b)52

c)64

d)76 - Jun 16th 2013, 05:25 AMProve ItRe: Quadratic equation question
Is your question:

If $\displaystyle \displaystyle \begin{align*} x + \frac{1}{16}x = 1 \end{align*}$ then find the value of $\displaystyle \displaystyle \begin{align*} 64x^3 + \frac{1}{64}x^3 \end{align*}$

or

If $\displaystyle \displaystyle \begin{align*} x + \frac{1}{16x} = 1 \end{align*}$ then find the value of $\displaystyle \displaystyle \begin{align*} 64x^3 + \frac{1}{64x^3} \end{align*}$? - Jun 16th 2013, 10:42 AMsscgeekRe: Quadratic equation question
IF x+ 1/16x = 1 then find the value of 64x^3 + 1/64x^3 is the correct question x is with 16 not 1/16 and x^3 is with 64 and not with 1/64

- Jun 16th 2013, 10:47 AMPlatoRe: Quadratic equation question
- Jun 16th 2013, 11:02 AMsscgeekRe: Quadratic equation question
1/(16x)

- Jun 16th 2013, 12:22 PMPlatoRe: Quadratic equation question
$\displaystyle \\1 = {\left( {x + \frac{1}{{16x}}} \right)^3}\\1 = {x^3} + \frac{{3x}}{{16}} + \frac{3}{{{{16}^2}x}} + \frac{1}{{{{16}^3}{x^3}}} \\1= \left( {{x^3} + \frac{1}{{{{16}^2}{x^3}}}} \right) + \frac{3}{{16}} \left( {x + \frac{1}{{16x}}} \right)$

Thus $\displaystyle \left( {{x^3} + \frac{1}{{{{16}^2}{x^3}}}} \right) = \frac{{13}}{{16}}$.

You finish. - Jun 16th 2013, 09:32 PMibduttRe: Quadratic equation question
I would request you to post the question again with sufficient parenthesis to avoid confusion and better understanding.