How to Prove this? If $a + b + c = 0$ $a + b + c = 0$ , Then Show that, $a^{2} - b c = b^{2} - c a = c^{2} - a b$ $a^2 - bc = b^2 - ca = c^2 - ab$ .

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Answer 1 · 2018-04-03T12:59:20

See the proof below

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

Then,

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

Therefore,

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

$= b^{2} + c^{2} + 2 b c - b c$ $=b^2+c^2+2bc-bc$

$= b^{2} + c^{2} + b c$ $=b^2+c^2+bc$

$b^{2} - c a = b^{2} - c (- (b + c)) = b^{2} + c^{2} + b c$ $b^2-ca=b^2-c(-(b+c))=b^2+c^2+bc$

$c^{2} - a b = c^{2} - (- (b + c) b) = c^{2} + b^{2} + b c$ $c^2-ab=c^2-(-(b+c)b)=c^2+b^2+bc$

So,

$a^{2} - b c = b^{2} - c a = c^{2} - a b$ $a^2-bc=b^2-ca=c^2-ab$

$Q E D$ $QED$

How to Prove this? If a + b + c = 0a+b+c=0a + b + c = 0, Then Show that, a^2 - bc = b^2 - ca = c^2 - aba2−bc=b2−ca=c2−aba^2 - bc = b^2 - ca = c^2 - ab.