Question

If $a^3+b^3=(8-3a-3b)ab$ and show that $\log \frac{a+b}{2}=\frac{1}{3}(\log a+\log b)$

Answer 1

$a^3+b^3=(8-3a-3b)ab$

$a^3+b^3=8ab-3a^{2}b-3a{b}^2$

$a^3+b^3+3a^{2}b+3a{b}^2=8ab$

$\frac{(a+b{)}^3}{2}=\log ab$

$\log \left(\frac{a+b}{2}\right)=\frac{1}{3}\log (a+b)$