Question

prove that: (a+b)3+(b+c)3+(c+a)3-3(a+b)(b+c)(c+a)=2(a3+b3+c3-3abc)

Answer 1

YOUR ANSWER IS IN THE ABOVE ATTACHMENT..⬇⬆

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$<b><i><u>FORMULAS WHICH IS USED :-$

$\boxed{ (x + y) {}^{3} = x {}^{3} + 3x {}^{2} y + 3xy {}^{2} + y {}^{3}}$



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✝✝…HOPE…IT…HELPS…YOU…✝✝

Answer 2

 (a+b)*3+(b+c)*3+(c+a)*3-3*(a+b)*(b+c)*(c+a)-(2*(a^3+b^3+c^3-3*a*b*c))=0 

Step by step solution :

Step  1  :

Checking for a perfect cube :

 1.1    a3-3abc+b3+c3  is not a perfect cube 

Equation at the end of step  1  :

(((((a+b)•3)+((b+c)•3))+((a+c)•3))-(((3•(a+b))•(b+c))•(a+c)))-2•(a3-3abc+b3+c3) = 0

Step  2  :

Equation at the end of step  2  :

(((((a+b)•3)+((b+c)•3))+((a+c)•3))-((3•(a+b)•(b+c))•(a+c)))-2•(a3-3abc+b3+c3) = 0

Step  3  :

Equation at the end of step  3  :

(((((a+b)•3)+((b+c)•3))+((a+c)•3))-(3•(a+b)•(b+c)•(a+c)))-2•(a3-3abc+b3+c3) = 0

Step  4  :

Equation at the end of step  4  :

(((((a+b)•3)+((b+c)•3))+((a+c)•3))-3•(a+b)•(b+c)•(a+c))-2•(a3-3abc+b3+c3) = 0

Step  5  :

Equation at the end of step  5  :

(((((a+b)•3)+((b+c)•3))+3•(a+c))-3•(a+b)•(b+c)•(a+c))-2•(a3-3abc+b3+c3) = 0

Step  6  :

Equation at the end of step  6  :

(((((a+b)•3)+3•(b+c))+3•(a+c))-3•(a+b)•(b+c)•(a+c))-2•(a3-3abc+b3+c3) = 0

Step  7  :

Equation at the end of step  7  :

(((3•(a+b)+3•(b+c))+3•(a+c))-3•(a+b)•(b+c)•(a+c))-2•(a3-3abc+b3+c3) = 0

Step  8  :

Step  9  :

Pulling out like terms :

 9.1     Pull out like factors :

   -2a3 - 3a2b - 3a2c - 3ab2 - 3ac2 + 6a - 2b3 - 3b2c - 3bc2 + 6b - 2c3 + 6c  = 

  -1 • (2a3 + 3a2b + 3a2c + 3ab2 + 3ac2 - 6a + 2b3+ 3b2c + 3bc2 - 6b + 2c3 - 6c) 

Equation at the end of step  9  :

-2a3 - 3a2b - 3a2c - 3ab2 - 3ac2 + 6a - 2b3 - 3b2c - 3bc2 + 6b - 2c3 + 6c = 0

Step  10  :

Solving a Single Variable Equation :

 10.1     Solve   -2a3-3a2b-3a2c-3ab2-3ac2+6a-2b3-3b2c-3bc2+6b-2c3+6c  = 0