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

Guest Jan 14, 2016

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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
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 #1
avatar+18712 
+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

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