without using the tables, find the value of (2+√3)^6 +(2-√3)^6

Answer 1

I see the sum of cubes
(2+√3)^6 +(2-√3)^6
= [(2+√3)^2]^3 + [(2-√3)^2]^3
= (4 + 4√3 + 3)^3 + (4 - 4√3) + 3)^3
= (7 + 4√3)^3 + (7 - 4√3)^3 , now from a^3 + b^3= (a+b)(a^2 - ab + b^2)
= (7+4√3 + 7-4√3)( (2+√3)^4 - (2+√3)^2 (2-√3)^2 + (2-√3)^4 )

look at the middle term, (2+√3)^2 (2-√3)^2
[(2+√3)(2-√3)]^2 = [4 - 3]^2 = 1 , nice!
for the first and last terms, expand it using the binomial expansion, things will cancel

= 14(.....) , believe it or not, that last part comes out to a whole number, I will let you work it out

let me know what your final answer is, I have written mine down