The Powers of Sum: Solving Hard Problem Algebraically

One of the active members of a Facebook group called Elite Math Circle (EMC) posted a question that seems interesting. Same question I enc0unter when I was actively solving problems at Brilliant.

It was posted by Russelle Guadalupe, an elite member of Philippine Team in 2011 International Mathematical Olympiad. The question is,

Given:

$a+b+c=6$
$a^2+b^2+c^2=8$
$a^3+b^3+c^3=5$

What is the value of $a^4+b^4+c^4$.

I have seen an elegant solution to this method, one of the members says it was “Newton’s Sum Theorem”.

The solution here might be the old school way to solve the problem using algebraic identities and manipulation.

Take note of the following identities:

1. $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$

2. $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$

First step is to find an expression that will lead to the unknown which is $a^4+b^4+c^4$. This expression is achievable by squaring $a^2+b^2+c^2$.

$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

But  $a^2+b^2+c^2=8$

$(8)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

$64=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

We still have one missing expression, $a^2b^2+b^2c^2+a^2c^2$.

Since we don’t have the latter expression, we will find it. Take a look of the first algebraic identity mentioned above. We can use it to arrive to an expression that we are looking for by squaring  $ab+bc+ac$.

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)$

We know that a+b+c=6,

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(6)$

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+12abc$ ♥♥

We already have an expression for our unknown but we ended now with two more unknowns. $ab+bc+ac$ and $abc$.

By squaring  $a+b+c$ and solving for  $ab+bc+ac$ we have,

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$

$(6)^2=8+2(ab+bc+ac)$

$2(ab+bc+ac)=36-8$

$ab+bc+ac=14$ ♥♥♥

We only have one unknown left, the $abc$

Recall the second identity mentioned above,

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$

or

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac))+3abc$

By substitution we have,

$5=(6)(8-14)+3abc$

$5=-36+3abc$

$41=3abc$

$164=12abc$ ♥♥♥♥

Substituting ♥♥♥ and ♥♥♥♥ to ♥♥ we have,

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+12abc$

$(14)^2=a^2b^2+b^2c^2+a^2c^2+164$

$196=a^2b^2+b^2c^2+a^2c^2+164$

$a^2b^2+b^2c^2+a^2c^2=196-164$

$a^2b^2+b^2c^2+a^2c^2=32$ ♥♥♥♥♥

Substituting ♥♥♥♥♥ to ♥ we have,

$64=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

$64=a^4+b^4+c^4+2(32)$

$64=a^4+b^4+c^4+64$

$\boxed{a^4+b^4+c^4=0}$

The method above has actually a routine, it has a pattern. In Mathematics, when there is a pattern, there is a general rule and formula can be derived. Dr. Greenie of Ask Dr. Math derived a formula using the following rules.

Given:

$a+b+c=x$
$a^2+b^2+c^2=y$
$a^3+b^3+c^3=z$

We can solve the value of $a^4+b^4+c^4$ in terms of x,y, and z with the following formula,

$a^4+b^4+c^4=y^2-2((\displaystyle\frac{x^2-y}{2})^2-\displaystyle\frac{x^4-3x^2y+2xz}{3})$

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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