+0  
 
0
794
4
avatar+262 

 Jun 8, 2015

Best Answer 

 #3
avatar+118587 
+10

$$\begin{array}{rlll}
\frac{3}{x^3-12x^2+48x-64}+\frac{111}{64-x^3}&=&0\qquad&(1)\\\\
\frac{3}{x^3-3(4x^2)+3(4^2x)-4^3}+\frac{111}{-(x^3-64)}&=&0\qquad&(2)\\\\
\frac{3}{(x-4)^3}-\frac{111}{x^3-64}&=&0\qquad&(3)\\\\
\frac{3}{(x-4)^3}-\frac{111}{(x-4)(x^2+4x+16)}&=&0\qquad&(4)\\\\
\frac{3(x^2+4x+16)}{(x-4)^3(x^2+4x+16)}-\frac{111(x-4)^2}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(5)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111(x^2-8x+16)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(6)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111x^2-8*111x+16*111}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(7)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}+\frac{-111x^2+888x-1776}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(8)\\\\
\frac{-108x^2+900x-1728}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(9)\\\\
\frac{-36(3x^2-25x+48)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(10)\\\\
x&=&\frac{25\pm 7}{6}\qquad \qquad&(11)\\\\
\end{array}$$

 

(11) came from the quadratic formula

NOW which line do you need explained?

 Jun 10, 2015
 #1
avatar+118587 
+5

(4-x)=-1(-4+x)=-(x-4)

 

Is that your only problem here?

 Jun 9, 2015
 #2
avatar+262 
+5

no i dont understand it thats exactly what i tried to do

 Jun 9, 2015
 #3
avatar+118587 
+10
Best Answer

$$\begin{array}{rlll}
\frac{3}{x^3-12x^2+48x-64}+\frac{111}{64-x^3}&=&0\qquad&(1)\\\\
\frac{3}{x^3-3(4x^2)+3(4^2x)-4^3}+\frac{111}{-(x^3-64)}&=&0\qquad&(2)\\\\
\frac{3}{(x-4)^3}-\frac{111}{x^3-64}&=&0\qquad&(3)\\\\
\frac{3}{(x-4)^3}-\frac{111}{(x-4)(x^2+4x+16)}&=&0\qquad&(4)\\\\
\frac{3(x^2+4x+16)}{(x-4)^3(x^2+4x+16)}-\frac{111(x-4)^2}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(5)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111(x^2-8x+16)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(6)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}-\frac{111x^2-8*111x+16*111}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(7)\\\\
\frac{3x^2+12x+48}{(x-4)^3(x^2+4x+16)}+\frac{-111x^2+888x-1776}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(8)\\\\
\frac{-108x^2+900x-1728}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(9)\\\\
\frac{-36(3x^2-25x+48)}{(x-4)^3(x^2+4x+16)}&=&0\qquad&(10)\\\\
x&=&\frac{25\pm 7}{6}\qquad \qquad&(11)\\\\
\end{array}$$

 

(11) came from the quadratic formula

NOW which line do you need explained?

Melody Jun 10, 2015
 #4
avatar+262 
+5

Thank you melody I understand everything now

 Jun 20, 2015

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