Foro de Matemáticas

kinu · Registrado: 04 Jul 2011 Mensajes: 5

If $a,b,c \in\mathbb{R}$ and $a+b+c=0\;\;,a^2+b^2+c^2=1$

then a^4+b^4+c^4=

kinu · Registrado: 04 Jul 2011 Mensajes: 5

$\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)$

Now Calculate value of $\left(a^2b^2+b^2c^2+c^2a^2\right)$

$\left(a+b+c\right)^2=a^2+b^2+c^2+2(ab+bc+ca)$

2(ab+bc+ca)=-1

$4.\left(a^2b^2+b^2c^2+c^2a^2\right)+2abc.\left(a+b+c\right)=-1$

So $\left(a^2b^2+b^2c^2+c^2a^2\right)=\displaystyle \frac{1}{4}$

So $\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)$

So $\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+\displaystyle 2.\frac{1}{4}$

So $\left(a^4+b^4+c^4\right)=\displaystyle\frac{1}{2}$