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!:)
$$\small{\text{$
12345678^4-(12345679^2+1)(12345677^2+1)
$}}$$
$$\small{\text{
We set $x = 12345678 $, so we have $x^4 -[ (x+1)^2+1 ]\cdot
[ (x-1)^2 +1 ]
$}}$$
$$\small{\text{$
\begin{array}{lcl}
x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] \\
&= & x^4 -[ x^2+2x+1+1 ]\cdot [ x^2-2x+1+1 ] \\
&= & x^4 -[ x^2+2x+2 ]\cdot [ x^2-2x+2 ] \\
&= & x^4 -[ (x^2+2)+2x ]\cdot [ (x^2+2)-2x ] \\
&= & x^4 -[ (x^2+2)^2 -(2x)^2 ] \\
&= & x^4 -[ x^4 +4x^2 + 4 -4x^2 ] \\
&= & x^4 -[ x^4 + 4 ] \\
&= & x^4 -x^4 - 4 \\
\mathbf{ x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ]}& \mathbf{= } & \mathbf{ - 4 } \\
\end{array}
$}}$$
.
$$\small{\text{$
12345678^4-(12345679^2+1)(12345677^2+1)
$}}$$
$$\small{\text{
We set $x = 12345678 $, so we have $x^4 -[ (x+1)^2+1 ]\cdot
[ (x-1)^2 +1 ]
$}}$$
$$\small{\text{$
\begin{array}{lcl}
x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ] \\
&= & x^4 -[ x^2+2x+1+1 ]\cdot [ x^2-2x+1+1 ] \\
&= & x^4 -[ x^2+2x+2 ]\cdot [ x^2-2x+2 ] \\
&= & x^4 -[ (x^2+2)+2x ]\cdot [ (x^2+2)-2x ] \\
&= & x^4 -[ (x^2+2)^2 -(2x)^2 ] \\
&= & x^4 -[ x^4 +4x^2 + 4 -4x^2 ] \\
&= & x^4 -[ x^4 + 4 ] \\
&= & x^4 -x^4 - 4 \\
\mathbf{ x^4 -[ (x+1)^2+1 ]\cdot [ (x-1)^2 +1 ]}& \mathbf{= } & \mathbf{ - 4 } \\
\end{array}
$}}$$