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Question : If $a^3=20a^2+b^2+c^2-a-340$, $b^3=20b^2+c^2+a^2-b-340$, $a^3=20a^2+a^2+b^2-c-340$, what is value of $abc$?

I think I should add this and get close to $abc$, but I can't think about this. I know that answer is 19. Please help me

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    @Should it be $c^{3}=20c^{2}+a^{2}+b^{2}-c-340$ in the third line? – Ishraaq Parvez Mar 21 '21 at 03:50
  • The answer cannot be unique: if we assume $a=b=c$ we see $a$ is a solution to $x^3-22x^2+x+340=0$, there are three real solutions, each one cubed gives a possibility of $abc$. – Yuval Mar 21 '21 at 04:23

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$$20a^2+b^2+c^2-a-340-a^3=0$$ $$20b^2+c^2+a^2-b-340-b^3=0$$ $$20c^2+a^2+b^2-c-340-c^3=0$$ Taking note of @Ishraaqparvez comments

Maybe before you find the product of $abc$, you should first know what their corresponding values are $$(a-19)^4(a^2+1)^4(a^3-22a^2+a+340)(a^4+5a^2-38a-356)(a^4-41a^3+422a^2+283a-6459)^2=0$$ $$(b-19)^4(b^2+1)^4(b^3-22b^2+b+340)(b^4+5b^2-38b-356)(b^4-41b^3+422b^2+283b-6459)^2=0$$ $$(c-19)^4(c^2+1)^4(c^3-22c^2+c+340)(c^4+5c^2-38c-356)(c^4-41c^3+422c^2+283c-6459)^2=0$$ Note that the equations are symmetric, so $a,b,c$ are different solutions of the same equation $$(x-19)(x^2+1)$$ $$x^3-22x^2+x+340$$ Then their real product is $-340$

Aderinsola Joshua
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