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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!:)

Guest Aug 19, 2015

Best Answer 

 #1
avatar+20009 
+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
 #1
avatar+20009 
+8
Best Answer

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

 

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

 

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

 

heureka  Aug 19, 2015
 #2
avatar+93342 
0

That is cool Heureka  

Melody  Aug 19, 2015

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