Question

(2+root(3))^6+[(2-root(3)]
without using calcultor

Answer 1

let
a=2+sqrt(3) b=2-sqrt(3)
but
a=((2+sqrt(3)^2)
b=[(2-sqrt(3)^2]
a^2=(2+sqrt(3))^2
=(2+sqrt(3)(2+sqrt(3)=7+4sqrt(3)
b^2=(2-sqrt(2)]^2
=(2-sqrt(3))(2-sqrt(3)
=7-4sqrt(3)
but
(a^3+b^3)=(a+b)(a^2-ab+b^2)
a+b=7+4sqrt(3)+7-4sqrt(3)
a+b=14
ab=(7-4sqr(3)(7+4sqrt(3)
ab=1
a^2=(7+4sqrt(3)(7+4sqrt(3)
=97+56sqrt(3)
b^2=(7-4sqrt(3)^2
b^2=(7-4sqrt(3)(7-4sqrt(3)=97-56sqrt(3)
a^3+b^3=(a+b)(a^2-ab+b^2)
=14(97+56sqrt(3)-1+96-56sqrt(3)
=14(97-1+97)
=14(193)=2702 yeah