Factorise (a + b + c)3 – (a3 + b3 + c3 )
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Answer:
Hope it's help full for u
Step-by-step explanation:
(a+b+c)
3
−a
3
−b
3
−c
3
=[(a+b+c)
3
−a
3
]−[b
3
+c
3
]
={(a+b+c−a)[(a+b+c)
2
+a
2
+(a+b+c)a]}−[(b+c)(b
2
+c
2
−bc)]
={(b+c)[(a+b+c)
2
+a
2
+(a+b+c)a]}−[(b+c)(b
2
+c
2
−bc)]
=(b+c)[(a+b+c)
2
+a
2
+a
2
+ab+ac−b
2
−c
2
+bc] =(b+c)[(a
2
+b
2
+c
2
+2ab+2bc+2ac)+2a
2
+ab+ac−b
2
−c
2
+bc]
=(b+c)[3a
2
+3ab+3bc+3ac]
=3(b+c)[a
2
+ab+bc+ac]
=3(b+c)[a(a+b)+c(b+a)]
=3(b+c)(a+b)(a+c)
Hence, (a+b+c)
3
−a
3
−b
3
−c
3
=3(a+b)(b+c)(c+a)
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