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2019 MathCon Finals Grade 11 Part C

I've been struggling with this one for quite a while, and I couldn't find any answers on the internet. If anyone can help, it would be greatly appreciated. Thank you

RandyMC
  • 11
  • From the two given equations, you can eliminate one variable and get an equation for the other alone. For example, eliminating $a$ you get $0=3b^9-4b^6+1=(b-1)(b^2+b+1)(3b^6-b^3-1)$. The vanishing of the factor $b^2+b+1$ doesn't produce real solutions. The factor $b-1$ gives $b=1$. We could compute the real solutions of $3b^6-b^3-1$, but we will only need $b^3$ which is one of the values $\frac{1\pm\sqrt{13}}{6}$. Since $a^3+b^3=(b^{-3})^2+b^3$. put there $b^3$ equal those values and multiply. – NDB Jul 12 '23 at 01:53
  • @NDB Sorry to trouble you again. How do comments work? says comments are not for answering a question. In case you do not agree with that established rule, please raise a question on meta.stackexchange. – Apass.Jack Jul 12 '23 at 02:04

1 Answers1

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First, let's rearrange the first equation for $a$. $$ab^2=1$$ $$a=\frac{1}{b^2}$$ Substituting this into the second equation, $$a^3+3b^3=4$$ $$\left(\frac{1}{b^2}\right)^3+3b^3=4$$ Let's let $x=b^3$. We can rewrite the previous equation as $$\frac{1}{x^2}+3x=4$$ $$1+3x^3=4x^2$$ $$3x^3-4x^2+1=0$$ You can now find the solutions to this equation. We can easily see that $x=1$ is a solution, so we can reduce this to a quadratic with polynomial long division. $$\left(x-1\right)\left(3x^2-x-1\right)=0$$ Using the quadratic formula, we find the solutions to be $$x=1,\frac{1+\sqrt{13}}{6},\frac{1-\sqrt{13}}{6}$$ The final part is to find the product of the possible values of $a^3+b^3$. Since $a=\frac{1}{b^2}$, this can be rewritten as $$\frac{1}{b^6}+b^3$$ Substituting $x$ from before, this is equivalent to $$\frac{1}{x^2}+x$$ However, this can also be rewritten as $(4-3x)+x=4-2x$ from an earlier identity. This means that the product will be equal to $$(4-2x_1)(4-2x_2)(4-2x_3)$$ Substituting the three solutions into this expression leaves you with the final answer.

riles
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  • +1, though, I'd suggest, instead of calculating the roots of $3x^3-4x^2+1$, we can observe that $\left(4-2 x_1\right) \left(4-2 x_2\right) \left(4-2 x_3\right)=-32\left(x_1+x_2+x_3\right)+16 \left(x_1 x_2+x_3 x_2+x_1 x_3\right)-8 x_1 x_2 x_3 +64$ and use Vieta's formulas. – Ennar Jul 12 '23 at 02:19
  • @Ennar, its simpler, as $\prod (a^3+b^3) = \prod (4-2b^3)=8\prod (2-x_i) = \dfrac83(3\cdot2^3-4\cdot2^2+1)=24$ – Macavity Jul 12 '23 at 04:51
  • @Macavity, indeed it is :) – Ennar Jul 12 '23 at 09:14