Ok. I really try to use the relation that since $A_{nxn}\quad$, $\det A^k = k^n \det A$
I can also use the $(a^3+b^3)$ and $(a^3-b^3)$ and stuck this expression seems is too much work for solve this problem.
$$[8^3-7^3-6\cdot 7^3 -4(7^3+6^3)+(5^3+4^3)+7\cdot 3^3+4(3^3-2^3)+1]\det A+1$$
Seems all the expression multiplier $\det A$ must be zero. But i suppose just the answer key.
By the Cayley-Hamilton theorem, $A$ is a root of its characteristic polynomial. Such polynomial is $$p_A(\lambda)=-\lambda^3+11\lambda^2+4\lambda-1\;,$$ so your matrix $B$ can be written as $$B=-A^5p_A(A)-Ap_A(A)+I=I\;.$$ Therefore $\det B=1$.
How did I notice? Well, in general, one feasible strategy could have been the following: we calculate the eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$ of $A$ (which are all real as $A$ is symmetric) and then we get the eigenvalues of $B=q(A)$ by applying $q$, so $q(\lambda_1)$, $q(\lambda_2)$, $q(\lambda_3)$. The determinant of $D$ is then the product of these three quantities.
This is true because, if $A$ is similar to the matrix $D$ by means of a base change $M$, then $A^k$ is similar to $D^k$ by means of the same base change, so $q(A)$ is similar to $q(D)$, for every polynomial $q$ with real coefficients.
However, as soon as I computed the characteristic polynomial of $A$ (in order to get the eigenvalues), I immediately noticed that our polynomial $q$ could be written as $-x^5p_A(x)-xp_A(x)+1$.