+0  
 
0
357
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+18956 
+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
Sort: 

1+0 Answers

 #1
avatar+18956 
+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

12 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details