Prove that

$\frac{59 \times 59 \times 59-9 \times 9 \times 9}{59 \times 59+59 \times 9+9 \times 9}$

$=\frac{(59)^{3}-9^{3}}{59^{2}+59 \times 9+9^{2}}$

We know

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

Here $a=59, b=9$

So $, \frac{(59-9)\left(59^{2}+9^{2}+59 \times 9\right)}{59^{2}+9^{2}+59 \times 9}=59-9=50: \mathrm{RHS}$

Thus, LHS=RHS