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((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

Guest Feb 26, 2017
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 #1
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((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

 

\(\large \frac{(a-1)^4-(b+1)^4}{(a-1)^3-(b+1)^3}=\frac{[(a-1)^2+(b+1)^2]\times [(a-1)^2-(b+1)^2]}{[(a-1)+(b+1)]\times [(a-1)-(b+1)]}\)

 

\(\large= \frac{(a^2-2a+1+b^2+2b+1)\times(a^2-2a+1-b^2-2b-1)}{(a+b)\times(a-b-2)}\)

 

\(\large= \frac{(a^2+2(b-a+1)+b^2)\times (a^2-2(a+b)-b^2)}{a^2-ab-2a+ab-b^2-2b}\)

 

\(\large= \frac{(a^2+2(b-a+1)+b^2)\times (a^2-2(a+b)-b^2)}{a^2-2(a+b)-b^2} \)

 

\(\large =a^2+2(b-a+1)+b^2\)    \(=a^2+2b-2a+2+b^2\)

 

laugh  !

asinus  Feb 26, 2017
edited by asinus  Feb 26, 2017
edited by asinus  Feb 26, 2017
 #3
avatar+6705 
0

((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

 

\(\large \frac{(a-1)^4-(b+1)^4}{(a-1)^3-(b+1)^3}\color{black}=\frac{[(a-1)^2+(b+1)^2]\times [(a-1)^2-(b+1)^2]}{a^3-3a^2+3a-b^3-3b^2-3b-2}\)

 

\(\large= \frac{(a^2-2a+1+b^2+2b+1)\times(a^2-2a+1-b^2-2b-1)}{a^3-3a^2+3a-b^3-3b^2-3b-2}\)

 

\(\large =\frac{a^4 - 4 a^3 + 6 a^2 - 4 a - b^4 - 4 b^3 - 6 b^2 - 4 b}{a^3-3a^2+3a-b^3-3b^2-3b-2}\)

 

\( Whoever \ can \ do \ it, \ makes \ something \ of \ it.\)

 

laugh  !

asinus  Feb 27, 2017
 #4
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((a - 1)^4 - (b + 1)^4)/((a - 1)^3 - (b + 1)^3)

 

=(a^3 + a^2 b - 2 a^2 + a b^2 + 2 a + b^3 + 2 b^2 + 2 b) / (a^2 + a b - a + b^2 + b + 1)

Guest Feb 27, 2017

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