2007 iTest Problems/Problem 39

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Problem

Let a and b be relatively prime positive integers such that a/b is the sum of the real solutions to the equation $\sqrt[3]{3x-4}+\sqrt[3]{5x-6}=\sqrt[3]{x-2}+\sqrt[3]{7x-8}$. Find $a+b$.

Solution

Let $A = \sqrt[3]{3x-4}, B = \sqrt[3]{5x-6}, C = \sqrt[3]{x-2},$ and $D = \sqrt[3]{7x-8}.$ With these substitutions, we know that \begin{align*} A+B &= C+D \\ A^3 + B^3 &= C^3 + D^3. \end{align*} We can factor the equation with the cubics and do some substitution. \begin{align*} (A+B)(A^2 - AB + B^2) &= (C+D)(C^2 - CD + D^2) \\ (A+B)((A+B)^2 - 3AB) &= (C+D)((C+D)^2 - 3CD) \end{align*} Bringing all the terms into one side results in \begin{align*} 0 &= (C+D)((C+D)^2 - 3CD) - (A+B)((A+B)^2 - 3AB) \\ 0 &= (A+B)((A+B)^2 - 3CD) - (A+B)((A+B)^2 - 3AB) \\ 0 &= (A+B)(3AB - 3CD) \end{align*} By the Zero Product Property, either $A+B = 0$ or $3AB-3CD = 0.$


If $A+B=0,$ then $\sqrt[3]{3x-4} = -\sqrt[3]{5x-6}.$ Cubing both sides yields $3x-4 = -5x+6,$ so $x = \tfrac54.$


If $3AB-3CD = 0,$ then $3\sqrt[3]{(3x-4)(5x-6)} = 3\sqrt[3]{(x-2)(7x-8)}.$ Dividing both sides by three and cubing both sides yields $(3x-4)(5x-6) = (x-2)(7x-8).$ Multiplying the binomials results in $15x^2 - 38x + 24 = 7x^2 - 22x + 16,$ and rearranging and factoring leads to $8(x-1)^2 = 0.$. That means $x=1.$


Since both values of $x$ satisfy the original equation, $\tfrac{a}{b} = \tfrac54 + 1 = \tfrac94,$ so $a+b = \boxed{13}.$

See Also

2007 iTest (Problems)
Preceded by:
Problem 38
Followed by:
Problem 40
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