Question

if a=2, b=1, c=3 then justify (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Answer 1

(a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc

So let’s write “a^2 + b^2 + c^2 - 2ab - 2bc - 2ca + 3abc” as:-

a^2+b^2+c^2–2ab-2ca+2bc-4bc+3abc

a^2+b^2+c^2–2ab-2ca+2bc-4bc+3abc

= (a-b-c)2 -4bc+3abc

= (a-b-c)2 –c(4b-3a)

Now putting values a = 2, b = 3, and c = (-2),

(2-3-(-2))2 – (-2)[4*3-3*2]

=(2-3+2)2 –(-2)(12-6)

=1-(-2)*6

=1-(-12)

=1+12

=13

Hope it helps...