prove that (a+b+c)^3-a^3-b^3-c^3=3 (a+b)(b+c)(c+a).
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Answer:
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Explanation:
(a+b+c)
3
=[(a+b)+c]
3
=(a+b)
3
+3(a+b)
2
c+3(a+b)c
2
+c
3
=>(a+b+c)
3
=(a
3
+3a
2
b+3ab
2
+b
3
)+3(a
2
+2ab+b
2
)c+3(a+b)c
2
+c
3
=>(a+b+c)
3
=a
3
+b
3
+c
3
+3a
2
b+3a
2
c+3ab
2
+3b
2
c+3ac
2
+3bc
2
+6abc
=>(a+b+c)
3
=a
3
+b
3
+c
3
+3a
2
b+3a
2
c+3ab
2
+3b
2
c+3ac
2
+3bc
2
+3abc+3abc
=>(a+b+c)
3
=a
3
+b
3
+c
3
+3a(ab+ac+b
2
+bc)+3c(ab+ac+b
2
+bc)
=>(a+b+c)
3
=a
3
+b
3
+c
3
+3(a+c)(ab+ac+b
2
+bc)
=>(a+b+c)
3
=a
3
+b
3
+c
3
+3(a+c)[a(b+c)+b(b+c)]
=>(a+b+c)
3
=a
3
+b
3
+c
3
+3(a+c)(b+c)(a+b)
=>(a+b+c)
3
−a
3
−b
3
−c
3
=3(a+c)(b+c)(a+b)
Hence,proved.
mark as brainliest answer.
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