Question

The sum of the reciprocals of two real numbers is 1 and the sum of their cubes is 4 then the numbers are

Answer 1

Answer:

Let the numbers be x and y

1/x+1/y =-1 , implying x+y=-xy. ……..(1)

We know that (x+y)^3= x^3+y^3+3xy(x+y)

Substitute from (1) and given sum of cubes,

-x^3y^3=4+3xy(-xy) = 4–3x^2y^2

Regrouping, equation becomes

X^2y^2(3-xy)=4

Factorise 4 ,we get factors as 1,2,2

This means xy = +2 or -2 and 3-xy=1 implying xy= 2 [ condition 3-xy not fulfilled with xy= -2]

Now this means x and y can be (1,2),(-1,-2)

But when we substitute this values in (1) the condition is false

So such a combination is not possible