Question

What is the value of $\frac{3^2+4^2+5^2+24+40+30}{3+4+5}$?

Answer 1

$3^2+4^2+5^2+24+40+30$ can be written as
$3^2+4^2+5^2+(2\times 3\times 4)+(2\times 4\times 5)+(2\times 5\times 3)$

Comparing the above equation with the identity
$(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$,
we get, $(3+4+5{)}^2=3^2+4^2+5^2+(2\times 3\times 4)+(2\times 4\times 5)+(2\times 5\times 3)$.
Hence,
$\frac{3^2+4^2+5^2+24+40+30}{3+4+5}=\frac{(3+4+5{)}^2}{(3+4+5)}=(3+4+5)=12$