3a2-b2-c2+2ab-2bc+2ca
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Answered by
1
Answer:
Step-by-step explanation:
3a2-b2-c2+2ab-2bc+2ca
3a²−b²−c²+2ab−2bc+2ac
=4a²−a²−b²−c²+2ab−2bc+2ac
=4a²−(a²+b²+c²−2ab+2bc−2ac)
=(2a)²−(a−b−c)²
=(2a+a−b−c)(2a−a+b+c)
=(3a−b−c)(a+b+c)
Answered by
1
Answer:
3a² - b² - c² + 2ab - 2bc + 2ca
= 4a² - a² - b² - c² + 2ab - 2bc + 2ca
= 4a² - (a² + b² + c² - 2ab + 2bc - 2ca)
= 4a² - (a - b - c)²
= (2a)² - (a - b - c)²
= (2a + a - b - c)(2a - a + b + c)
= (3a - b - c)(a + b + c)
hope it helps you!!
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