prove that a*sq+b*sq.+c*sq.+2ab+2bc+2ca=a+b+c whole square
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Answered by
1
1.(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
LHS:
(a+b+c)² = (a+b+c) × (a+b+c)
= a²+ab+ca+ab+b²+bc+ca+bc+c²
= a²+b²+c²+2ab+2bc+2ca
RHS:
a²+b²+c²+2ab+2bc+2ca
Therefore,
LHS = RHS
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
LHS:
(a+b+c)² = (a+b+c) × (a+b+c)
= a²+ab+ca+ab+b²+bc+ca+bc+c²
= a²+b²+c²+2ab+2bc+2ca
RHS:
a²+b²+c²+2ab+2bc+2ca
Therefore,
LHS = RHS
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
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Answered by
4
Step-by-step explanation:
1.(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
lhs:
(a+b+c)² = (a+b+c) × (a+b+c)
= a²+ab+ca+ab+b²+bc+ca+bc+c²
= a²+b²+c²+2ab+2bc+2ca
rhs:
a²+b²+c²+2ab+2bc+2ca
therefore,
lhs = rhs
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca.
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