I have two questions. First, is it true to say that $$(((x^m + y^n)^2)+z)^0 = 1$$ for when $m$ is the index of $x$ and $n$ is the index of $y$, and when $z$ is a natural number? Second question, could I potentially solve this equation for a variable, as the first step would be me taking the zeroth root of the equation. Is it possible?
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Since $u^0 = 1$ for any $u \neq 0$, we cannot determine much about $x, y, z$ from this equation. – Travis Willse Apr 06 '17 at 15:02
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Relevant: http://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1. – Martín-Blas Pérez Pinilla Apr 07 '17 at 07:34
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$(((x^m + y^n)^2)+z)^0 =\dfrac{(((x^m + y^n)^2)+z)^k}{(((x^m + y^n)^2)+z)^k }=1$
But notice that $0^0=\dfrac{0}{0}$ which is not defined. Therefore$t^0=1$ is only true when $t\neq 0$
Fawad
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The equation is satisfied by all $x,y,z$ if you know that $z$ is a natural number (assuming $m,n>0$, and assuming these are exponents -- you called them "indices").
That's because $(x^m+y^n)^2\geq0$, and $z\geq 1$, so the expression carrying the exponent $0$ is a strictly positive number.
If you don't know that $m,n>0$, then the expression $x^m+y^n$ may not be defined for nonpositive $x,y$.
MPW
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