Solve that a²-b²-c²+2bc,b²-c²-a²+2ac,c²-a²-b²+2ab
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Step-by-step explanation:
Given : a
2
−b
2
+2bc−c
2
We can write it as
a
2
−b
2
+2bc−c
2
=a
2
−(b
2
−2bc+c
2
)
We know that, (a−b)
2
=a
2
−2ab+b
2
a
2
−(b
2
−2bc+c
2
)=a
2
−(b−c)
2
Based on the identity, a
2
−b
2
=(a+b)(a−b)
We get,
a
2
−(b−c)
2
=[a+(b−c)[a−(b−c)]
⟹a
2
−(b−c)
2
=(a+b−c)(a−b+c)
∴ a
2
−b
2
+2bc−c
2
=(a+b−c)(a−b+c)
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