# solving for x.....

• Mar 22nd 2008, 06:45 PM
pakman134
solving for x.....
I know this sounds a little stupid and it may sound like I am just trying to get answers but really clueless. The only thing I could really think of is factoring out (d-3)^3 and then finding a lowest common denominator so I can multiply by the bottom and get rid of the fraction and then factor out the x's and solve for them... But I keep getting lost in all my work... Would appreciate any help. Thanks!

Solve for x.

$
0=\frac{-2c}{x^3}+\frac{6c}{(d-x)^3}
$
• Mar 22nd 2008, 06:50 PM
topsquark
Quote:

Originally Posted by pakman134
I know this sounds a little stupid and it may sound like I am just trying to get answers but really clueless. The only thing I could really think of is factoring out (d-3)^3 and then finding a lowest common denominator so I can multiply by the bottom and get rid of the fraction and then factor out the x's and solve for them... But I keep getting lost in all my work... Would appreciate any help. Thanks!

Solve for x.

$
0=\frac{-2c}{x^3}+\frac{6c}{(d-x)^3}
$

$0 = -\frac{2c}{x^3} + \frac{6c}{(d - x)^3}$

Multiply both sides by $x^3(d - x)^3$:
$0 = -2c(d - x)^3 + 6cx^3$

Now expand:
$0 = -2cd^3 + 6cd^2x - 6cdx^2 + 2cx^3 + 6cx^3$

$8cx^3 - 6cdx^2 + 6cd^2x - 2cd^3 = 0$

$4x^3 - 3dx^2 + 3d^2x - d^3 = 0$

I see no way to continue from here without a MAJOR operation involving something like Cardano's method.

-Dan
• Mar 22nd 2008, 07:01 PM
bobak
First combine the fractions

$\frac{6c}{(d-x)^3}- \frac{2c}{x^3} \Rightarrow \frac{6cx^3 - 2c(d-x)^3}{(d-x)^3x^3}$

so we have $\frac{6cx^3 - 2c(d-x)^3}{(d-x)^3x^3} = 0$

therefore $6cx^3 - 2c(d-x)^3 = 0$
$\Rightarrow 3x^3 - (d-x)^3 = 0$
$\Rightarrow (3^{\frac{1}{3}}x)^3 - (d-x)^3 = 0$

now you use the difference of two cubes

$(3^{\frac{1}{3}}x -(d-x))((3^{\frac{1}{3}}x)^2 + 3^{\frac{1}{3}}x(d-x) + (d-x)^2 ) = 0$

deal with the linear factor to get $x = \frac{d}{3^{\frac{1}{3}} + 1}$