(a+b+c)2=a2+b2+c2+2ab+2b+2ca---verify
Anonymous:
with all the steps plz
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Answered by
44
(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
=(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
Is it your project?
means my frnd gave me
its as imp as pro so plz
then let me add some more steps...
ok
ok now?
its enough if u can add more its better
thank u i will surely mark as best
:P
thank alot thank u
Answered by
11
(a+b+c)²
let a+b=x ,c=y
substitute the values in the equation
(x+y)²=x²+y²+2xy
(a+b)²+c²+2c(a+b)
a²+b²+2ab+c²+2ac+2bc
a²+b²+c²+2ab+2bc+2ca
let a+b=x ,c=y
substitute the values in the equation
(x+y)²=x²+y²+2xy
(a+b)²+c²+2c(a+b)
a²+b²+2ab+c²+2ac+2bc
a²+b²+c²+2ab+2bc+2ca
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