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

Answer 1

Hazel Anglin

See the proof below

If #a+b+c=0#

Next,

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

Consequently,

#a^2-bc=(-(b+c))^2-bc=(b+c)^2-bc#

#=b^2+c^2+2bc-bc#

#=b^2+c^2+bc#

#b^2-ca=b^2-c(-(b+c))=b^2+c^2+bc#

#c^2-ab=c^2-(-(b+c)b)=c^2+b^2+bc#

Thus,

#a^2-bc=b^2-ca=c^2-ab#

#QED#

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Answer 2

Genesis Allen

Start with the given equation: (a + b + c = 0). Square both sides of the equation. Expand the squared terms. Rearrange the terms to obtain the required expressions: (a^2 - bc = b^2 - ca = c^2 - ab).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.