if x = (sqrt 3 + 1)/2 then find x^3 + 1/x^3
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1/x=1/(sqrt3+1)=1/( sqrt3+1 x (sqrt3-1)/(sqrt3-1) = (sqrt3-1)/2
Cubing both the side we get
(x+1/x)^3 = x^3 + (1/x)^3 + 3(x + 1/x)
={sqrt3+1+(sqrt3-1)/2)}^3 = x^3 +(1/x)^3 + 3(sqrt3+1+(sqrt3-1)/2)
={(3sqrt3+1)/2}^3=x^3 + (1/x)^3 + 3{(3sqrt3+1)/2)
={(3sqrt3+1)/2}^3 - {3{(3sqrt3+1)/2)}
= (81sqrt3+1)/8-{(9sqrt3+3)/2}
here "-" sign will change signs after the bracket
taking lcm
(81sqrt3+1-36sqrt3-12)/8
=(45sqrt3-11)/8
Cubing both the side we get
(x+1/x)^3 = x^3 + (1/x)^3 + 3(x + 1/x)
={sqrt3+1+(sqrt3-1)/2)}^3 = x^3 +(1/x)^3 + 3(sqrt3+1+(sqrt3-1)/2)
={(3sqrt3+1)/2}^3=x^3 + (1/x)^3 + 3{(3sqrt3+1)/2)
={(3sqrt3+1)/2}^3 - {3{(3sqrt3+1)/2)}
= (81sqrt3+1)/8-{(9sqrt3+3)/2}
here "-" sign will change signs after the bracket
taking lcm
(81sqrt3+1-36sqrt3-12)/8
=(45sqrt3-11)/8
digvijayreddy:
x= (sqrt 3+1)/2 so 1/x becomes 2/(sqrt 3+1) but i understood the process. thanks in helping me
nope 1/x becomes (sqrt3-1)/2
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