Question

find the value of (ab+bc)(ab-bc)+(bc+ca)(bc-ca)+(ca+ab)(ca-ab)​

Answer 1

Answer:

(ab+bc)(ab-bc)+(bc+ca)(bc-ca)+(ca+ab)(ca-ab)=ab^2-bc^2+bc^2-ca^2+ca^2-ab^2=0

Answer 2

Step-by-step explanation:

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