BMAD 62 Report post Posted May 7, 2013 The sum of the reciprocals of two real numbers is −1, and the sum of their cubes is 4. What are the numbers? Share this post Link to post Share on other sites

0 witzar 18 Report post Posted May 7, 2013 I would go with binomial formula: (x+y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3} = (x^{3} + y^{3}) + 3xy(x+y) We know that x^{3} + y^{3} = 4 and 1/x + 1/y = -1 or x + y = -xy. Let's introduce new variable: z = x + y = -xy. We can now substitute binomial formula with: z^{3} = 4 - 3z^{2}, which is equivalent to (z-1)(z+2)^{2} = 0. Last thing to do is to solve system of equations: x + y = z xy = -z for z = 1 and for z = -2. For z=1 we get "golden ratio" solutions: (1 ± sqrt(5))/2, and z=-2 case has no real solutions. Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted May 7, 2013 Our old friend the golden ratio. Let: m = (x-y)/2 and n = (x+y)/2 1/m + 1/n = [2/(x-y) + 2/(x+y)] / (x^{2}-y^{2}) = -1 This simplifies toy^{2} = x^{2} + 4x Similarly, m^{3} + n^{3} = 4 expands and simplifies tox(x^{2} + 3y^{2}) = 16 Substitute y^{2} from previous result to getx^{2}(x+3) = 4 of which x=1 is clearly a root. Then from the above first result, y^{2}=5 givesm = (1 - sqrt(5))/2 = -0.618033989 n = (1 + sqrt(5))/2 = 1.618033989 = phi = 1-m now it happens that n = -1/m so that the obvious m+n = 1 leads directly to1/m + 1/n = -1m^{3} = -0.2360679775n^{3}. = 4.2360679775 whencem^{3} + n^{3} = 4 Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted May 7, 2013 I would go with binomial formula: (x+y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3} = (x^{3} + y^{3}) + 3xy(x+y) We know that x^{3} + y^{3} = 4and 1/x + 1/y = -1or x + y = -xy. Let's introduce new variable: z = x + y = -xy.We can now substitute binomial formula with: z^{3} = 4 - 3z^{2},which is equivalent to (z-1)(z+2)^{2} = 0. Last thing to do is to solve system of equations: x + y = z xy = -zfor z = 1 and for z = -2. For z=1 we get "golden ratio" solutions: (1 ± sqrt(5))/2,and z=-2 case has no real solutions. Very nice! Share this post Link to post Share on other sites

The sum of the reciprocals of two real numbers is −1, and the sum of their cubes is 4. What are the numbers?

## Share this post

## Link to post

## Share on other sites