If a = 12, b = 13, c = 15, then the value of a³ + b³ + c³ – 3abc is ;

a = 12, b = 13, c = 15

$\Rightarrow a+b+c=12+13+15=40$

$\therefore a^{3}+b^{3}+c^{3}-3 a b c$

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

$=\frac{1}{2}(a+b+c)$

$\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$

$=\frac{1}{2} \times 40$

$=\left[(12-13)^{2}+(13-15)^{2}+(15-12)^{2}\right]$

$=20\left(1+2^{2}+3^{2}\right)$

$=20 \times(1+4+9)$

$=20 \times 14=280$