Prove that

To Prove:

$(x+1+i)(x+1-i)(x-1+i)(x-1-i)=\left(x^{4}+4\right)$

Taking LHS

$(x+1+i)(x+1-i)(x-1+i)(x-1-i)$

$=[(x+1)+i][(x+1)-i][(x-1)+i][(x-1)-i]$

Using $(a-b)(a+b)=a^{2}-b^{2}$

$=\left[(x+1)^{2}-(i)^{2}\right]\left[(x-1)^{2}-(i)^{2}\right]$

$=\left[x^{2}+1+2 x-i^{2}\right]\left(x^{2}+1-2 x-i^{2}\right]$

$=\left[x^{2}+1+2 x-(-1)\right]\left(x^{2}+1-2 x-(-1)\right]\left[\because i^{2}=-1\right]$

$=\left[x^{2}+2+2 x\right]\left[x^{2}+2-2 x\right]$

Again, using $(a-b)(a+b)=a^{2}-b^{2}$

Now, $a=x^{2}+2$ and $b=2 x$

$=\left[\left(x^{2}+2\right)^{2}-(2 x)^{2}\right]$

$=\left[x^{4}+4+2\left(x^{2}\right)(2)-4 x^{2}\right]\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=\left[x^{4}+4+4 x^{2}-4 x^{2}\right]$

$=x^{4}+4$

$=R H S$

$\therefore$ LHS = RHS

Hence Proved