newton in maths!!! how

Algebra Level 3

-127 120 147 110 138

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1 solution

Chew-Seong Cheong
May 22, 2015

Using Newston's Sums method, we have:

{ r + s + t = 3 r s + s t + t r = 4 r s t = 8 \begin{cases} r+s+t & = -3 \\ rs+st+tr & = 4 \\ rst & = 8 \end{cases}

And:

r 2 + s 2 + t 2 = ( r + s + t ) 2 2 ( r s + s t + t r ) = ( 3 ) 2 2 ( 4 ) = 1 r 3 + s 3 + t 3 = ( r + s + t ) ( r 2 + s 2 + t 2 ) ( r s + s t + t r ) ( r + s + t ) + 3 r s t = ( 3 ) ( 1 ) 4 ( 3 ) + 3 ( 8 ) = 33 r 4 + s 4 + t 4 = ( r + s + t ) ( r 3 + s 3 + t 3 ) ( r s + s t + t r ) ( r 2 + s 2 + t 2 ) + r s t ( r + s + t ) = ( 3 ) ( 33 ) 4 ( 1 ) + 8 ( 3 ) = 127 \begin{aligned} r^2 + s^2 + t^2 & = (r+s+t)^2 - 2(rs+st+tr) = (-3)^2 -2(4) = 1 \\ r^3 + s^3 + t^3 & = (r+s+t)(r^2 + s^2 + t^2) - (rs+st+tr)(r+s+t) \\ & \quad \quad + 3rst \\ & = (-3)(1)-4(-3)+3(8) = 33 \\ r^4 + s^4 + t^4 & = (r+s+t)(r^3 + s^3 + t^3) - (rs+st+tr)(r^2 + s^2 + t^2) \\ & \quad \quad + rst(r+s+t) \\ & = (-3)(33)-4(1)+8(-3) = \boxed{-127} \end{aligned}

Are Vieta's sum something different from this?

Abhijeet Verma - 5 years, 11 months ago

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No, they are the same

Chew-Seong Cheong - 5 years, 11 months ago

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