+0

# ((a−1) 4 −(b+1) 4 )/((a−1) 3 −(b+1) 3 )

+5
274
4

((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

Guest Feb 26, 2017
#1
+7348
+5

((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

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

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

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

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

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

!

asinus  Feb 26, 2017
edited by asinus  Feb 26, 2017
edited by asinus  Feb 26, 2017
#3
+7348
0

((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

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

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

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

$$Whoever \ can \ do \ it, \ makes \ something \ of \ it.$$

!

asinus  Feb 27, 2017
#4
0

((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