Recognising sqrt(3)/2 being cos(theta) and 1/2 being sin(theta), (sqrt(2)/2 + 1/2) ^9 can be reduced to ((cos(theta) + sin(theta))^2)^4 * (cos(theta) + sin(theta)) (cos(theta) + sin(theta))^2 = 1 + sin(2. theta) So the expression is boiled down to (1 + sin(2*theta))^4 * (sin(theta) + cos(theta)) The power 4 term can be easily expanded by the Binomial's theorem. Then, the expanded terns can be further multiplied by (cos(theta) + sin(theta)) The solution is then arrived. My 2 cents
Recognising sqrt(3)/2 being cos(theta) and 1/2 being sin(theta),
(sqrt(2)/2 + 1/2) ^9 can be reduced to
((cos(theta) + sin(theta))^2)^4 * (cos(theta) + sin(theta))
(cos(theta) + sin(theta))^2 = 1 + sin(2. theta)
So the expression is boiled down to
(1 + sin(2*theta))^4 * (sin(theta) + cos(theta))
The power 4 term can be easily expanded by the Binomial's theorem. Then, the expanded terns can be further multiplied by (cos(theta) + sin(theta))
The solution is then arrived.
My 2 cents