Can You Prove Gy^3+(y-G)^3 Equals Zero Given x^2+x+G=0?

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The discussion focuses on proving the equation Gy^3 + (y - G)^3 = 0 given the quadratic equation x^2 + x + G = 0 and the expression y = x + 1/x. Participants analyze the derived expressions for Gy^3 and (y - G)^3, concluding that they are not equal. A key point of confusion arises regarding the correct interpretation of y, with clarification confirming that y = x + 1/x is the accurate formulation. The conversation emphasizes the importance of correctly applying algebraic identities in the proof.
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Homework Statement



If x^2+x+G=0 and y=x+1/x , prove that Gy^3+(y-G)^3=0

Homework Equations





The Attempt at a Solution



Gy^3=x^5+3x^3+3x+1/x+x^4+3x^2+3+1/x^2

(y-G)^3=1/x^3+6/x+3+12x+12x^2+12x^4+6x^5+x^6

They are not equal.

Is this the correct way of doing ?
 
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Does
y = x + \frac{1}{x}
OR
y = \frac{x + 1}{x}
?


69
 
eumyang said:
Does
y = x + \frac{1}{x}
OR
y = \frac{x + 1}{x}
?


69

y=x+\frac{1}{x}
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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