An algebra problem by Sheikh Sakib Ishrak Shoumo

Algebra Level 3

Let the zeros of 7 x 2 + x + 7 = 0 7x^2+x+7=0 be α \alpha and β \beta . Now, α 3 + β 3 = p q \alpha^3 + \beta ^3 =\dfrac pq for some positive coprime integers p p and q q . Find p + q p+q .


The answer is 489.

Sheikh Sakib Ishrak Shoumo by Sheikh Sakib Ishrak Shou… , 22

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1 solution

Using Vieta's formula , we get, α + β = − 1 7 \alpha+\beta=-\frac 17 ; and, α β = 1 \alpha \beta=1 .

Thus, α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β ) = ( − 1 7 ) 3 − 3 ( 1 ) ( − 1 7 ) = 146 343 \alpha^3 +\beta^3= (\alpha+\beta)^3-3\alpha \beta ( \alpha+\beta)=(-\frac 17)^3-3(1)(-\frac 17)=\frac {146}{343}

And so, the answer is 489 \boxed{489} .

N.B.: We may fall in trouble if we try using the complex roots here, in lieu of Vieta's.

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