SOLUTION: I need detail solution: A = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} + {{{sqrt(5)}}})( {{{sqrt(2)}}} − {{{sqrt(3)}}} + {{{sqrt(5)}}}) and B = ( {{{sqrt(2)}}} + {{{sqrt(3)}}} −

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Question 824392: I need detail solution:
A = ( sqrt%282%29 + sqrt%283%29 + sqrt%285%29)( sqrt%282%29 − sqrt%283%29 + sqrt%285%29) and B = ( sqrt%282%29 + sqrt%283%29 − sqrt%285%29)(−sqrt+%282%29 + sqrt%283%29 + sqrt%285%29) find C=AB
(In the "A" I've got 4-2sqrt%2815%29, but I can't solve B)
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
A) let u = square root (2) + square root (5)
then we can rewrite problem as
(u + square root(3)) * (u - square root(3))
u^2 - 3
(2 + 2*square root(2)*square root(5) +5) - 3
(2 +2*square root(10) +5 -3)
(4 +2*square root(10))
2*(2 + square root(10))
B) let v = square root(2) - square root (5)
then we can rewrite problem as
(square root(3) + v) * (square root(3) - v)
3 - v^2
3 - (2 -2*square root(2)*square root(5)+5)
3 - (2 -2*square root(10)+5)
3 - 2 +2*square root(10) - 5
2*square root(10) - 4
2*(square root(10) - 2)
now we are asked what is C = AB
C = (2*square root(10) + 4) * (2*square root(10) -4)
C = 4*10 - 16
C = 40 -16
C = 24