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If the real number $a,b,c$ satisfy the equation $$\frac{3x^3-2x^2+x+1}{3x^3-2x^2-x-1}=\frac{3x^3-2x^2+5x-13}{3x^3-2x^2-5x+13}$$ then find the value of $18(a+b+c)$.

I got a solution as $7÷2$ as letting $3x^3-2x^2=m$ and $x+1=n$ and $5x-13=l$ and by applying componento and dividendo we get $n= l$ which gives $x=7÷2$ but I don't know the other solutions.Please help.

Parcly Taxel
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Rajendra Sharma
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1 Answers1

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You have approached this the right way, but you neglected one possibility.

Multiplying out by the denominators, your equation becomes $$m^2+(n-l)m-ln=m^2-(n-l)m-ln$$ This gives you either $n-l=0$ (which you have found) or $m=0$.

Do remember, when you find $x$ and substitute it back into the original equation, to check that neither of the denominators becomes zero.

Martin Kochanski
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