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#1

$Old$ September 30th, 2009, 08:27 AM

anshulbshah $anshulbshah is offline$

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$Post$ A olympiad question

If p and q are the roots of the equation 3x^2 + x- 1 = 0 then prove that 3(p^3 + q^3) + (p^2 +q^2) - (p + q) = 0

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$Old$ September 30th, 2009, 08:43 AM

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Originally Posted by anshulbshah $View Post$

If p and q are the roots of the equation 3x^2 + x- 1 = 0 then prove that 3(p^3 + q^3) + (p^2 +q^2) - (p + q) = 0

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$3x^2+x-1=0$

$x^2+\frac{1}{3}x-\frac{1}{3}=0$

$p+q=-\frac{1}{3}$ , $pq=-\frac{1}{3}$

$p^3+q^3=(p+q)(p^2+q^2-pq)$

$=(p+q)[(p+q)^2-3pq]$

$=(-\frac{1}{3})[(-\frac{1}{3})^2-3(-\frac{1}{3})]$

Then

$p^2+q^2=(p+q)^2-2pq$

$=(-\frac{1}{3})^2-2(-\frac{1}{3})$

Then putting them together would give you 0 .

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