Find $a^{b^{c}}$ given that $\frac{1}{2}{a^2}+b^{2}+c^{2}-bc-ac=2(b-1).$

Question

Find $\space\large{a^{b^{c}}}$ given that $\dfrac{1}{2}{a^2}+b^{2}+c^{2}-bc-ac=2(b-1).$

Hi everyone, I attempted to apply the $(a+b+c)^{2}$ result to solve this to no avail.

For advice, please!

Answer 1

Think of squares, you can get the answer.

$0=a^2/2+b^2+c^2-bc-ac-2(b-1)=(a-c)^2/2+(b-c)^2/2+(b-2)^2/2$. So you can get $a=b=c=2$ thus $a^{b^c}=16$