+0  
 
0
448
1
avatar

how do you get rid of the bottom? [(x^3-2x)-(a^3-2a)]/(x-a)

Guest Jan 14, 2016

Best Answer 

 #1
avatar+19653 
+15

how do you get rid of the bottom? [(x^3-2x)-(a^3-2a)]/(x-a)

 

\(\begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-2x-a^3+2a } {x-a}\\ &=& \frac{ x^3-a^3-2x+2a } {x-a}\\ &=& \frac{ x^3-a^3-2(x-a) } {x-a}\\ &=& \frac{ x^3-a^3 }{x-a} - 2\cdot( \frac{ x-a } {x-a} ) \\ &=& \frac{ x^3-a^3 }{x-a} - 2\\ \end{array}\\ \boxed{~ \text{Difference of two cubes:}\\ \begin{array}{rcl} x^3-a^3 &=& (x-a)(x^2+x\cdot a+a^2) \end{array} ~}\\ \begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-a^3 }{x-a} - 2\\ &=&\frac{ (x-a)(x^2+x\cdot a+a^2) }{x-a} - 2\\ &=& x^2+x\cdot a+a^2 - 2\\ \end{array} \)

 

laugh

heureka  Jan 14, 2016
 #1
avatar+19653 
+15
Best Answer

how do you get rid of the bottom? [(x^3-2x)-(a^3-2a)]/(x-a)

 

\(\begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-2x-a^3+2a } {x-a}\\ &=& \frac{ x^3-a^3-2x+2a } {x-a}\\ &=& \frac{ x^3-a^3-2(x-a) } {x-a}\\ &=& \frac{ x^3-a^3 }{x-a} - 2\cdot( \frac{ x-a } {x-a} ) \\ &=& \frac{ x^3-a^3 }{x-a} - 2\\ \end{array}\\ \boxed{~ \text{Difference of two cubes:}\\ \begin{array}{rcl} x^3-a^3 &=& (x-a)(x^2+x\cdot a+a^2) \end{array} ~}\\ \begin{array}{rcl} \frac{ (x^3-2x)-(a^3-2a) } {x-a} &=& \frac{ x^3-a^3 }{x-a} - 2\\ &=&\frac{ (x-a)(x^2+x\cdot a+a^2) }{x-a} - 2\\ &=& x^2+x\cdot a+a^2 - 2\\ \end{array} \)

 

laugh

heureka  Jan 14, 2016

5 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.