Simple, fun algebra challenge

The above equation can be simplified into:

$\frac{A^{3}}{8} = \frac{A^{2}}{6} \cdot C = A^2$

which $C = 6$ and $A^3 - 8A^2 = 0 \Rightarrow A= 0,8.$ But, $A = 0$ leads to division-by-zero (not allowed). So, $\boxed{A = 8}$ is the answer.

$\left( \frac { A }{ 2 } \right) \left( \frac { A }{ 2 } \right) \left( \frac { A }{ 2 } \right) =A\cdot A\\ \frac { { A }^{ 3 } }{ 8 } ={ A }^{ 2 }\\ \frac { A }{ 8 } =1\\ A=8$

Solution- So to solve for C we discover that (A(squared)/2)/3 x C= A squared based on that we know that 2 x 3=C as (A(squared)/2)/3 x C= Equals the exact same thing the equation started with. So C=6 Now lets solve for A (A/2)(cubed)= A(squared), whelp the answer is 8!!! as we divide 8 into 2 we get 4. 4(squared)= 8(squared) divided by 4, This occurs with all numbers (check the equation below). So based on that 8 is the only number that has 8/2 (cubed)=8(squared),

equation X= b/f, B(squared)=f(squared) times x(squared), X(squared)=B(squared)/ f (squared) .