If a=2, b=3 and c=4 then find the numerical value of a2+2abc+b2+c2

a=2 b=3 and c=4. then,

a2+2abc+b2+c2

a2+2abc+b2+c2= 2^2+2×2×3×4+3^2+4^2

(replace value of a, b, c by2,3,4 respectively tgen solve)

= 4+48+9+16

= 77………………

Therefore, 77 is correct answer.

it would open into a square depending on the type of envelope i’m not sure about the second part