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

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

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Reagan 2 weeks 2021-09-11T11:50:58+00:00 1 Answer 0 views 0

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    2021-09-11T11:52:50+00:00

    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

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