Reciprocals and Cubes

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Posted · Report post

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

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Posted · Report post

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

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Posted · Report post

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.

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Posted · Report post

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!

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