Is this equation on exponentiation true?

Question

Is the below equation true for all real values of $a$, $b$, $c$ and $d$? (with $c\neq0$)

$$d(a+b)^c= (ad^{1/c}+bd^{1/c})^c$$

Some cases has been tested and the above has not been disproved.

Answer 1

It fails for $c=1/2$, $d=-1$, $a,b=1$, since then $$ d(a+b)^c=-2^{1/2}=-\sqrt2\neq\sqrt2=(ad^{1/c}+bd^{1/c})^c=2^{1/2}$$ However, it is valid (and both sides are defined) whenever the bases of exponentiation are positive: when $d>0$ and $a+b>0$. This is because $ab^x=a^xb^x$ and $(a^x)^y=a^{xy}$ are valid for real $a,b>0$ and real $x,y$, so that the RHS can then be rewritten $$(ad^{1/c}+bd^{1/c})^c=(d^{1/c}(a+b))^c=(d^{1/c})^c(a+b)^c=d^1(a+b)^c=d(a+b)^c.$$