Reciprocals and Cubes

[BMAD]

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

[bonanova]

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)] / (x²-y²) = -1

This simplifies to
y² = x² + 4x

Similarly, m³ + n³ = 4 expands and simplifies to
x(x² + 3y²) = 16

Substitute y² from previous result to get
x²(x+3) = 4 of which x=1 is clearly a root.

Then from the above first result, y²=5 gives

m = (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 to
1/m + 1/n = -1

m³ = -0.2360679775
n³. = 4.2360679775

whence
m³ + n³ = 4

[witzar]

I would go with binomial formula:

(x+y)³ = x³ + 3x²y + 3xy² + y³ = (x³ + y³) + 3xy(x+y)

We know that

x³ + y³ = 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:

z³ = 4 - 3z²,

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.

[bonanova]

I would go with binomial formula: (x+y)

³ = x³ + 3x²y + 3xy² + y³ = (x³ + y³) + 3xy(x+y) We know that x³ + y³ = 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: z³ = 4 - 3z²,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!