Let p , q and r be distinct real roots of the equation x 3 − 4 x 2 − 6 x + 1 = 0 . Find the numerical value of the expression:
( q − p ) ( p − r ) p 5 + ( r − q ) ( q − p ) q 5 + ( p − r ) ( r − q ) r 5
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@Kobe Mercado , I have found a simpler solution.
Easier solution
From x 3 − 4 x 2 − 6 x + 1 = 0 , we have:
x 3 x 4 x 5 = 4 x 2 + 6 x − 1 = 4 x 3 + 6 x 2 − x = 4 ( 4 x 2 + 6 x − 1 ) + 6 x 2 − x = 2 2 x 2 + 2 3 x − 4 = 2 2 x 3 + 2 3 x 2 − 4 x = 2 2 ( 4 x 2 + 6 x − 1 ) + 2 3 x 2 − 4 x = 1 1 1 x 2 + 1 2 8 x − 2 2
Since p , q , and r are the roots, they all satisfy the equations above. Then:
X = ( p − r ) ( q − p ) p 5 + ( q − p ) ( r − q ) q 5 + ( r − q ) ( p − r ) r 5 = ( q − p ) ( r − q ) ( p − r ) p 5 ( r − q ) + q 5 ( p − r ) + r 5 ( q − p ) = p 2 q + q 2 r + r 2 p − p q 2 − q r 2 − r p 2 1 1 1 ( r p 2 − p 2 q + p q 2 − q 2 r + q r 2 − r 2 p ) + 1 2 8 ( r p − p q + q r − r p + q r − r p ) − 2 2 ( r − p + p − r + q − p ) = p 2 q + q 2 r + r 2 p − p q 2 − q r 2 − r p 2 1 1 1 ( p q 2 + q r 2 + r p 2 − p 2 q − q 2 r − r 2 p ) + 1 2 8 ( 0 ) − 2 2 ( 0 ) = − 1 1 1
Similar solution as @Kobe Mercado's
Note that by Vieta's formula , we have p + q + r = 4 , p q + q r + r p = − 6 , and p q r = − 1 .
X = ( p − r ) ( q − p ) p 5 + ( q − p ) ( r − q ) q 5 + ( r − q ) ( p − r ) r 5 = ( q − p ) ( r − q ) ( p − r ) p 5 ( r − q ) + q 5 ( p − r ) + r 5 ( q − p ) = ( q − p ) ( r − q ) ( p − r ) − ( q − p ) ( r − q ) ( p − r ) ( p 3 + q 3 + r 3 + p 2 q + p 2 r + q 2 p + q 2 r + p r 2 + q r 2 + p q r ) = − ( p 3 + q 3 + r 3 + p 2 q + p 2 r + q 2 p + q 2 r + p r 2 + q r 2 + p q r ) = − ( ( p + q + r ) ( ( p + q + r ) 2 − 3 ( p q + q r + r p ) ) + 3 p q r + p q ( p + q ) + q r ( q + r ) + r p ( r + p ) + p q r ) = − ( ( p + q + r ) ( ( p + q + r ) 2 − 3 ( p q + q r + r p ) ) + ( p q + q r + r p ) ( p + q + r ) + p q r ) = − 4 ( 1 6 + 1 8 ) − ( − 6 ) ( 4 ) + 1 = − 1 1 1
You forgot the negative sign after you cancelled out the common factors in the third line of your solution because r-q and q-r are additive inverses.
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Combining those 3 fractional expressions will yield:
( q − p ) ( p − r ) ( r − q ) p 5 ( r − q ) + q 5 ( p − r ) + r 5 ( q − p ) .
The numerator of the resulting fraction can be factored into ( q − p ) ( p − r ) ( q − r ) ( p ³ + q ³ + r ³ + p ² q + p ² r + q ² p + q ² r + p r ² + q r ² + p q r ) .
Upon simplifying, the expression is equivalent to − ( p ³ + q ³ + r ³ + p ² q + p ² r + q ² p + q ² r + p r ² + q r ² + p q r ) = − [ p ³ + q ³ + r ³ + p ² ( q + r ) + q ² ( p + r ) + r ² ( p + q ) + p q r ] .
By Vieta's Theorem, p + q + r = 4 ⟹ q + r = 4 − p . Similarly, p + r = 4 − q and p + q = 4 − r .
Therefore, \(-[p³+q³+r³+p²(q+r)+q²(p+r)+r²(p+q)+pqr]=-[p³+q³+r³+p²(4-p)+q²(4-q)+r²(4-r)+pqr]
=-[p³+q³+r³-p³-q³-r³+4(p²+q²+r²)+pqr]
=-4(p²+q²+r²)-pqr\).
By Vieta's Theorem, p + q + r = 4 and \(pq+qr+pr=-6 \implies p²+q²+r²=(p+q+r)²-2(pq+qr+pr)
=4²-2(-6)=
16+12=28\). In addition, by Vieta's Theorem, p q r = − 1 .
Thus, the expression is equal to:
− 4 ( 2 8 ) − ( − 1 ) = − 1 1 2 + 1 = − 1 1 1 .