• Jun 16th 2013, 04:45 AM
sscgeek
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 AM
Prove It

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 AM
sscgeek
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 AM
Plato
Quote:

Originally Posted by sscgeek
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

You still have not answered the question. Use parentheses.

Is it 1/(16x) or is it (1/16)x?
• Jun 16th 2013, 11:02 AM
sscgeek
1/(16x)
• Jun 16th 2013, 12:22 PM
Plato
Quote:

Originally Posted by sscgeek
If x + 1/(16x) = 1, then the value of 64 x^3 + 1/(64x^3) is

a)4 b)52 c)64 d)76

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

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

You finish.
• Jun 16th 2013, 09:32 PM
ibdutt