Question

please answer this question fully explain I mark you as a brainlist ok help me​

Answer 1

$a+b+c=0$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac) \\ \\ a^3+b^3+c^3-3abc = 0(a^2+b^2+c^2-ab-bc-ac) \\ \\ a^3+b^3+c^3-3abc = 0 \\ \\ a^3+b^3+c^3=3abc \\ \\ \frac{a^3+b^3+c^3}{abc}=3$

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$\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab} \\ \\ \\ \frac{a^2(ac)+b^2(bc)}{(bc)(ac)}+\frac{c^2}{ab} \\ \\ \\ \frac{a^3c+b^3c}{abc^2}+\frac{c^2}{ab} \\ \\ \\ \frac{(a^3c+b^3c)(ab)+c^2(abc^2)}{(abc^2)(ab)} \\ \\ \\ \frac{a^4bc+ab^4c+abc^4}{a^2b^2c^2} \\ \\ \\ \frac{abc(a^3+b^3+c^3)}{(abc)^2} \\ \\ \\ \frac{a^3+b^3+c^3}{abc}=3$

$\therefore\ \frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=3$

$Hence proved! \\ \\ \\ Thank you. :-)$

Answer 2

if a + b + c= 0

then

a^3 + b^3+ c^3 = 3abc

$\frac{a^3 + b^3+ c^3}{abc} = 3$

a^2/bc + b^2/ac + c^2/ab = 3