+0

# Algebraic Manipulation

0
655
2

Hi can anyone help me?

How to solve 12345678^4-(12345679^2+1)(12345677^2+1)?

The answer is -4 but whyy?? Thanks so much!:)

Aug 19, 2015

#1
+22869
+8

$$\small{\text{ 12345678^4-(12345679^2+1)(12345677^2+1) }}$$

$$\small{\text{ We set x = 12345678 , so we have x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] }}$$

$$\small{\text{ \begin{array}{lcl} x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] \\ &= & x^4 -[ x^2+2x+1+1 ]\cdot [ x^2-2x+1+1 ] \\ &= & x^4 -[ x^2+2x+2 ]\cdot [ x^2-2x+2 ] \\ &= & x^4 -[ (x^2+2)+2x ]\cdot [ (x^2+2)-2x ] \\ &= & x^4 -[ (x^2+2)^2 -(2x)^2 ] \\ &= & x^4 -[ x^4 +4x^2 + 4 -4x^2 ] \\ &= & x^4 -[ x^4 + 4 ] \\ &= & x^4 -x^4 - 4 \\ \mathbf{ x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ]}& \mathbf{= } & \mathbf{ - 4 } \\ \end{array} }}$$

Aug 19, 2015

#1
+22869
+8

$$\small{\text{ 12345678^4-(12345679^2+1)(12345677^2+1) }}$$

$$\small{\text{ We set x = 12345678 , so we have x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] }}$$

$$\small{\text{ \begin{array}{lcl} x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] \\ &= & x^4 -[ x^2+2x+1+1 ]\cdot [ x^2-2x+1+1 ] \\ &= & x^4 -[ x^2+2x+2 ]\cdot [ x^2-2x+2 ] \\ &= & x^4 -[ (x^2+2)+2x ]\cdot [ (x^2+2)-2x ] \\ &= & x^4 -[ (x^2+2)^2 -(2x)^2 ] \\ &= & x^4 -[ x^4 +4x^2 + 4 -4x^2 ] \\ &= & x^4 -[ x^4 + 4 ] \\ &= & x^4 -x^4 - 4 \\ \mathbf{ x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ]}& \mathbf{= } & \mathbf{ - 4 } \\ \end{array} }}$$

heureka Aug 19, 2015
#2
+102761
0

That is cool Heureka

Aug 19, 2015