Question

a²+b²+c²-x²+2ab+2bc+2ca

Answer 1

Step-by-step explanation:

(a+b+c)²

=(a+b+c) × (a+b+c)

=[a×(a+b+c)] + [b×(a+b+c)] +[c×(a+b+c)] (by distributive law)

= [a² + ab + ac] + [ ba +b² + bc] + [ ca + cb + c²] 

= a² + ab + ac + ba +b² + bc + ca + cb + c²

= a² + ab + ac + ab +b² + bc + ac + bc + c²   (ab=ba)

= a² + b² + c² + ab + ab + bc + bc + ac + ac

= a² + b² + c² + 2ab + 2bc + 2ca

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Answer 2

Answer:

(a+b+c)

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

(a+b+c+x)(a+b+c-x)