priscillianaw1

2022-09-30

If $a+b+c=0$ and ${a}^{2}+{b}^{2}+{c}^{2}=1$, work out ${a}^{4}+{b}^{4}+{c}^{4}$.

Could this problem admit a solution through the method of lagrange multipliers.

Could this problem admit a solution through the method of lagrange multipliers.

Zara Pratt

Beginner2022-10-01Added 12 answers

Let $a,b,c$ be the roots of a cubic polynomial.

Since $a+b+c=0$ this has form ${x}^{3}+px+q=0$ where

$$p=ab+bc+ca=\frac{{\textstyle (a+b+c{)}^{2}-({a}^{2}+{b}^{2}+{c}^{2})}}{{\textstyle 2}}=-\frac{{\textstyle 1}}{{\textstyle 2}}$$

Now note multiply by $x$ to see that $a,b,c$ satisfy $f(x)={x}^{4}+p{x}^{2}+qx=0$

Then

$$0=f(a)+f(b)+f(c)=({a}^{4}+{b}^{4}+{c}^{4})+p({a}^{2}+{b}^{2}+{c}^{2})+q(a+b+c)$$

Substituting known values gives ${a}^{4}+{b}^{4}+{c}^{4}=\frac{{\textstyle 1}}{{\textstyle 2}}$

Working with polynomials is sometimes a convenient way of capturing and organising information about symmetric functions.

Since $a+b+c=0$ this has form ${x}^{3}+px+q=0$ where

$$p=ab+bc+ca=\frac{{\textstyle (a+b+c{)}^{2}-({a}^{2}+{b}^{2}+{c}^{2})}}{{\textstyle 2}}=-\frac{{\textstyle 1}}{{\textstyle 2}}$$

Now note multiply by $x$ to see that $a,b,c$ satisfy $f(x)={x}^{4}+p{x}^{2}+qx=0$

Then

$$0=f(a)+f(b)+f(c)=({a}^{4}+{b}^{4}+{c}^{4})+p({a}^{2}+{b}^{2}+{c}^{2})+q(a+b+c)$$

Substituting known values gives ${a}^{4}+{b}^{4}+{c}^{4}=\frac{{\textstyle 1}}{{\textstyle 2}}$

Working with polynomials is sometimes a convenient way of capturing and organising information about symmetric functions.

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