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Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)?
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Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3a...
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Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3a...
Proof:
We need to prove the identity: (a + b)³ (b + c)³ (c + a)³ - 3(a + b)(b + c)(c + a) = 2(a³ + b³ + c³ - 3abc)
Expand the LHS:
(a + b)³ = a³ + 3a²b + 3ab² + b³
(b + c)³ = b³ + 3b²c + 3bc² + c³
(c + a)³ = c³ + 3c²a + 3ca² + a³
Now, we can expand the LHS as follows:
LHS = (a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³)
To simplify this expression, we need to expand each term in the brackets and combine like terms.
Expand the first term:
(a³ + 3a²b + 3ab² + b³) = a³ + 3a²b + 3ab² + b³
Expand the second term:
(b³ + 3b²c + 3bc² + c³) = b³ + 3b²c + 3bc² + c³
Expand the third term:
(c³ + 3c²a + 3ca² + a³) = c³ + 3c²a + 3ca² + a³
Multiply the expanded terms:
(a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³) = (a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³)
Expanding this expression will result in a very long and complex polynomial expression. However, we can simplify it by grouping like terms.
Simplify the expression:
(a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³) = (a³b³c³ + 3a³b³c²a + 3a³b³ac² + a³b³a³ + 3a³b²c³b + 9a³b²c²a² + 9a³b²cac² + 3a³b²a³b + 3a³bc³c + 9a³bc²a²c + 9a³bcac²c + 3a³bca³ + b³c³c³ + 3b³c³c²a + 3b³c³ac² + b³c³a³ + 3
We need to prove the identity: (a + b)³ (b + c)³ (c + a)³ - 3(a + b)(b + c)(c + a) = 2(a³ + b³ + c³ - 3abc)
Expand the LHS:
(a + b)³ = a³ + 3a²b + 3ab² + b³
(b + c)³ = b³ + 3b²c + 3bc² + c³
(c + a)³ = c³ + 3c²a + 3ca² + a³
Now, we can expand the LHS as follows:
LHS = (a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³)
To simplify this expression, we need to expand each term in the brackets and combine like terms.
Expand the first term:
(a³ + 3a²b + 3ab² + b³) = a³ + 3a²b + 3ab² + b³
Expand the second term:
(b³ + 3b²c + 3bc² + c³) = b³ + 3b²c + 3bc² + c³
Expand the third term:
(c³ + 3c²a + 3ca² + a³) = c³ + 3c²a + 3ca² + a³
Multiply the expanded terms:
(a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³) = (a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³)
Expanding this expression will result in a very long and complex polynomial expression. However, we can simplify it by grouping like terms.
Simplify the expression:
(a³ + 3a²b + 3ab² + b³)(b³ + 3b²c + 3bc² + c³)(c³ + 3c²a + 3ca² + a³) = (a³b³c³ + 3a³b³c²a + 3a³b³ac² + a³b³a³ + 3a³b²c³b + 9a³b²c²a² + 9a³b²cac² + 3a³b²a³b + 3a³bc³c + 9a³bc²a²c + 9a³bcac²c + 3a³bca³ + b³c³c³ + 3b³c³c²a + 3b³c³ac² + b³c³a³ + 3
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Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)?
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Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)?.
Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Prove that:(a +b)3 (b +c)3 (c +a)3-3(a +b)(b +c)(c +a)=2(a3 +b3+ c3-3abc)?.
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