Question

If $a+b+c=0$ then prove

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

Answer 1

Given: If $a+b+c=0$

We know that $(x+y+z{)}^2=x^2+y^2+z^2+2xy+2yz+2zx$

then $(a+b+c{)}^2=0$ [squaring both sides]

$a^2+b^2+c^2+2(ab+bc+ca)=0$

$a^2+b^2+c^2=-2ab-2bc-2ca$

On squaring on both sides, we get

L.H.S $=(a^2+b^2+c^2{)}^2=a^4+b^4+c^4+2a^2{b}^2+2b^2{c}^2+2c^2{a}^2$

R.H.S $=(-2ab-2bc-2ca{)}^2=4a^2{b}^2+4b^2{c}^2+4c^2{a}^2+8a{b}^{2}c+8b{c}^{2}a+8c{a}^{2}b$

$\therefore a^4+b^4+c^4+(2a^2{b}^2+2b^2{c}^2+2c^2{a}^2)=(4a^2{b}^2+4b^2{c}^2+4c^2{a}^2)+8a{b}^{2}c+8b{c}^{2}a+8c{a}^{2}b$

$\Rightarrow a^4+b^4+c^4=(4a^2{b}^2+4b^2{c}^2+4c^2{a}^2)-(2a^2{b}^2+2b^2{c}^2+2c^2{a}^2)+8a{b}^{2}c+8b{c}^{2}a+8c{a}^{2}b$

$\Rightarrow a^4+b^4+c^4=(2a^2{b}^2+2b^2{c}^2+2c^2{a}^2)+8a{b}^{2}c+8b{c}^{2}a+8c{a}^{2}b$

$\Rightarrow a^4+b^4+c^4=(2a^2{b}^2+2b^2{c}^2+2c^2{a}^2)+8abc(b+c+a)$

$\Rightarrow a^4+b^4+c^4=(2a^2{b}^2+2b^2{c}^2+2c^2{a}^2)+8abc(a+b+c)$

$\Rightarrow a^4+b^4+c^4=(2a^2{b}^2+2b^2{c}^2+2c^2{a}^2)+8abc\times 0$ [$\because (a+b+c)=0$]

$\Rightarrow a^4+b^4+c^4=2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$ [proved]