How can we prove that:
in the following equation?
$$ 3bx^{a+4} = -24x^6 $$
Can we not find any values of $b$ or $a$ that satisfy the above equation such that $3b$ does NOT equal to $-24$ and/or that $a+4$ does NOT equal to $6$?
On
Let $p(x) = a_0 + a_1 x + \dots + a_n x^n$ and $q(x) = b_0 + b_1 x + \dots + b_m x^m$ be two polynomials. Then, $p = q$ if and only if $m = n$ and $a_i = b_i$ for each $0 \leq i \leq n$.
In other words, two polynomials are equal if and only if they have the same degree and the coefficient of each monomial is the same in both polynomials.
Here, the LHS is the polynomial $3b x^{a+4}$ and the RHS is the polynomial $-24x^6$. By the above statements, these polynomials are equal if and only if $$a + 4 = 6 \quad \text{and} \quad 3b = -24.$$
As in my comment . . .
Presumably, in the context of the problem, $b$ is an unknown constant, and $x$ is an indeterminate, which can take any value.
So letting $x=1$, we get $3b=-24$, hence $b=-8$.
Replacing $3b$ by $-24$, the given identity reduces to $x^{a+4}=x^6$, and $x$ can still take any value.
So letting $x=2$, we get $2^{a+4}=2^6$, which implies $a+4=6$, hence $a=2$.
In this solution, we are not assuming $3bx^{a+2}$ is specified to be a polynomial.
But if the context of the problem allows that assumption, then as in Brahadeesh's answer, it's automatic that $a+4=6$, else the degrees won't match, and then we must have $3b=-24$, else the corresponding coefficients won't match.