#### Question

Let

a=\dfrac{1}{x}
(1)
b=\dfrac{1}{x+1}
(2)

The given equation is converted to an integer equation.

a^3-b^3=\dfrac{7}{8}
(3)

From the definition of a and b, take reciprocal of the equations (1) and (2)

\dfrac{1}{a}=x, \dfrac{1}{b}=x+1

Subtraction of them gives

\dfrac{1}{a}-\dfrac{1}{b}=-1

Rearrange the equation, we get

a-b=ab
(4)

Now Let's evaluate the cube of binomial formula

(a-b)^3=a^3+3ab^2-3a^2b-b^3

Rearrange terms of the right hand side

(a-b)^3=a^3-b^3-3ab(a-b)

Substituting (3) and (4) gives a cubic equation in terms of ab

\dfrac{7}{8}=(ab)^3-3(ab)^2

Let

ab = p
(5)
the equation is simplified as

8p^3+24p^2-7=0

Factor the terms in the left hand side,

8p^3-4p^2+28p^2-7=0

4p^2(2p-1)+7(4p^2-1)=0

4p^2(2p-1)+7(2p-1)(2p+1)=0

(2p-1)(4p^2+14p+7)=0

2p-1=0
(6)

or

4p^2+14p+7=0
(7)

Solving the linear equation, we get our first solution

p=\dfrac{1}{2}

Solving the quadratic equation (7), we get tow solutions

p=\dfrac{-14\pm\sqrt{14^2-4\times 4\times 7 } }{8}

=\dfrac{-7\pm\sqrt{42} }{4} <0

However, from equation (5), (4) and (1), (2)

p=ab =a-b=\dfrac{1}{x}-\dfrac{1}{x+1}

From the given condition

\dfrac{1}{x^3} -\dfrac{1}{(x+1)^3} =\dfrac{7}{8}>0

\dfrac{1}{x}-\dfrac{1}{x+1}>0

\therefore p>0

Therefore the two solutions with square root were rejected, Only solution is

p=ab=\dfrac{1}{2}

Plug in ab in (4), then

a-b = \dfrac{1}{2}

Substituting a and b with (1) and (2)

\dfrac{1}{x}-\dfrac{1}{x+1}= \dfrac{1}{2}

\dfrac{1}{x(x+1)}=\dfrac{1}{2}

x^2+x-2=0

Therefore

x=-2 or x=1

Steven Zheng posted 6 days ago

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