Quartic Sum Determination

Algebra Level 4

$\large x^3-8x^2+5x+12 = 0$

If $A,B$ and $C$ are roots of the equation above, find the value of $A^4 + B^4+C^4$ .

The answer is 2482.

2 solutions

Danish Ahmed
Aug 8, 2015

Equating coefficients in

$(x - A)(x - B)(x - C) = x^3 - 8x^2 + 5x + 12$ gives:

$A + B + C = 8$

$AB + BC + CA = 5$

$ABC = -12.$

Using these values in the following identity:

$A^4 + B^4 + C^4 = (A + B + C)^4 - 4(A + B + C)^2(AB + BC + CA) +2(AB + BC + CA)^2 + 4ABC(A + B + C)$

$A^4 + B^4 + C^4 = 84 - 4*82*5 + 2*52 - 4*12*8 = 2482$

Moderator note:

Your identity for $A^4+B^4+C^4$ is not commonly known. How did you come up with that?

Alternatively, there's a simpler approach.

Hint 1 : Newton's Sum. Or Newton's identity.

Hint 2 : $A^2 + B^2+ C^2 = (A+B+C)^2-2(AB+AC+BC)$ .

the solution comes from newton binomials, right??

WeiLiang Wu - 5 years, 10 months ago

Ben Habeahan
Sep 2, 2015

you can read newton identities

here.

Let $P_1=A+B+C, \\ P_2=AB+BC+CA , \\ P_3=ABC,$

with vieta's formula $P_1=8, P_2=5, P_3=-12.$

Using newton identities, we have: $S_1=P_1=8 \\ S_2=P_1S_1-2P_2=(8)(8)-2(5)=54 \\ S_3=P_1S_2-P_2S_1+3P_3=(8)(54)-(5)(8)+(3)(-12)=356 \\ S_4=P_1S_3-P_2S_2+P_3S_1=(8)(356)-(5)(54)+(-12)(8)= \boxed{2482}$

Quartic Sum Determination

The answer is 2482.

2 solutions

Moderator note:

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