Question

If $a+b+c=0$, and $x+y+z=0$, shew that $(a^3+b^3+c^3+d^3{)}^2=9(bcd+cda+dab+abc{)}^2=9(bc-ad)(ca-bd)(ab-cd)$.

Answer 1

$\Rightarrow \sum a^3=3r$ when $a+b+c+d=0$

$\therefore {(\sum a^3)}^2=9r^2=9{(\sum abc)}^2$

$=9{(abc+bcd+bad+cda)}^2$

Since, $d=-(a+b+c)$

$bcd+cda+dab+abc=-(a+b+c)(bc+ab+ca)+abc$

$=-(b+c)(c+a)(a+b)$

$\therefore {(bcd+cda+dab+abc)}^2={(b+c)}^2{(c+a)}^2{(a+b)}^2$

$(c+a)(a+b)=bc+a(a+b+c)$

$=bc-ad$

$\therefore 9{(\sum abc)}^2=9(bc-ad)(ca-bd)(ab-cd)$