Tricky Ratio Puzzle

This is an interesting algebra problem from BL’s Weekly Math Games, which is behind a subscription wall.

(a3 – b3)/(a – b)3 = 73/3,

what is a – b?”

In fact, it is possible to solve for a and b individually as well.

Answer.

See a Tricky Ratio Puzzle for the solution.

(Updates 11/6/2024, 11/9/2024)  Other Solutions

(Update 11/6/2024)  BL’s Solution

Rummaging around I found the beginnings of BL’s solution, again because I am not a subscriber.  I added the remaining steps as I imagined them.  His solution is simpler.

See BL’s Solution

(Update 11/9/2024)  Oscar Roja’s Solution

Oscar Rojas has a different take on a solution.

See Oscar Rojas’s Solution

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One thought on “Tricky Ratio Puzzle”

  1. Sanjay Godse

    LHS = (a^3-b^3)/(a-b)^3
    =(a^2 + a*b + b^2)/((a-b)^2
    =(a^2 – 2*a*b + b^2 + 3*a*b)/((a-b)^2
    =[ (a-b)^2 + 3*a*b]/(a-b)^2
    = 1 + 3*a*b/(a-b)^2 = RHS = 73/3
    3*a*b/(a-b)^2 = 73/3 -1 = 70/3
    a*b / (a-b)^2 = 70/9
    since 70 and 9 are relatively prime,
    a*b = 70 and a-b = sqrt(9) = 3
    a*b=70 can be expressed as
    70*1 or 35*2 or 14*5 or 10*7

    Only 10 and 7 gives a-b=10-7=3.

    Hence a=10 and b=7
    *************

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