Q.A zero of $x^3+64$ is :

Composing $f$ of degree $m$ with $g$ of degree $n$ produces a leading term $x^{mn}$, so the degree multiplies rather than adds.

1. Let $f(x)=a_mx^m+\cdots$ (leading term $a_mx^m$, $a_m\ne0$) and $g(x)=b_nx^n+\cdots$ (leading term $b_nx^n$, $b_n\ne0$).
2. $h(x)=(f\circ g)(x)=f(g(x))=a_m\big(g(x)\big)^m+\cdots$. …

The equation has one real root (the real cube root of $-64$) and two complex roots; checking each option against $x^3=-64$.

1. $x^3+64=0\Rightarrow x^3=-64$.
2. Testing $x=-4$: $(-4)^3=-64$. This satisfies the equation.
3. Testing $x=4i$: $(4i)^3=64i^3=64(-i)=-64i\ne-64$. Not a zero. …

The Fundamental Theorem of Algebra guarantees exactly $n$ roots (counted with multiplicity, in $\mathbb{C}$) for a degree-$n$ polynomial equation — not necessarily real or distinct.

1. A polynomial equation of degree $n$ has the form $a_0x^n+a_1x^{n-1}+\cdots+a_n=0$, with $a_0\ne0$.
2. The Fundamental Theorem of Algebra states that such an equation has at least one root in $\mathbb{C}$.
3. Applying this repeatedly (factoring out each root) shows the polynomial factors completely into $n$ linear factors over $\mathbb{C}$, so it has exactly $n$ roots when multiplicities are counted.
4. These $n$ roots need NOT all be distinct — a root can repeat (multiplicity $>1$), so "$n$ distinct roots" is not guaranteed. …