Let $A, B$ be $n\times n$ with $n\ge 2$ nonsingular matrices with real entries such that $A^{-1} + B^{-1} =(A+B)^{-1}$ 
Let $A, B$ be $n\times n$ with $n\ge 2$ nonsingular matrices with real entries such that $$A^{-1} + B^{-1} =(A+B)^{-1}$$
  then prove that $\operatorname{det}(A)=\operatorname{det}(B)$.
  Also show that this result is not valid for complex matrices.

I'm not getting any way out to solve this. Can anybody solve the problem?
Thanks for assistance in advance.
 A: By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain  $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$\det(U)=1$$or equivalently$$\det(A)=\det(B)$$

Counter example on $\Bbb C$
Let $A=I_2$  and $b=kI_2$ with $k={-1+i\sqrt 3\over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1\over k}I\\(A+B)^{-1}={1\over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$\det(B)=k^2=-1-k={-1-i\sqrt 3\over 2}\ne 1=\det(A)$$

Comment
Even on $\Bbb C$, from $U^3=I$ we can conclude that$$\left|\det(A)\right|=\left|\det(B)\right|$$
A: For $A,B \in \mathcal{M}_n^n(\mathbb{R})$
We've $$(A+B)(A+B)^{-1}=(A+B)(A^{-1}+B^{-1})=I+AB^{-1}+BA^{-1}+I=I\\
\implies AB^{-1}+BA^{-1}+I=O$$
Then post multiplying the equation , once by $B$ and then by $A$ we end up with
$$AB^{-1}A=BA^{-1}B=-(A+B)$$
Considering the determinants we get
$$\frac{[det(A)]^2}{det(B)}=\frac{[det(B)]^2}{det(A)} \implies [det(A)]^{3}-[det(B)]^{3}=0 \implies det(A)=det(B)$$
If we're working with $\mathcal{M}^n_n(\mathbb{C})$ , the result doesn't hold.
Let's define $\zeta_{n}^{j}$ as the $n$th roots of unity for some odd positive integer $n$ with $1 \le j \le n$ and $j \in \mathbb{Z}^{+}$, then let's define $A=\zeta_n I_{n}$ and $B=\sum_{j=2}^{n}\zeta_{n}^{j}I_n$
Clearly $$A^{-1} +B^{-1}=I\left[\zeta_{n}^{-1} +(\sum_{j=2}^{n}\zeta_{n}^{j})^{-1} \right]=O$$
and $$(A+B)^{-1}=O$$
But we note that
$$det(A)=(\zeta_{n})^n =1 \\
det(B)=(\sum_{j=2}^{n}\zeta_{n}^{j})^n=(-\zeta_{n})^n =-1$$
NOTE:Properties of $\zeta_n^{j}$ are used in the arguments without any mention
