Question

Prove that (X+y)cube + (y+z) cube+(z+X) cube- 3(X+y)(y+z)(z+X)=2(xcube+ycube+zcube-3xyz)

Answer 1

$\huge \boxed{ \mathbb{ \ulcorner ANSWER: \urcorner}}$


$\boxed{See \: the \: attachment \: for \: your \: answer.}$


$\huge \boxed{ \mathbb{ALGEBRAIC \: \: IDENTITIES}}$



$\huge \bf \underline{ \mathbb{LHS = RHS .}}$



✔✔ Hence, it is proved ✅✅.

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Answer 2

Heya!!

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Solve LHS:

(x+y) ^3 + (y+z) ^3 +(z+x) ^3 - 3(x+y)(y+z)(x+z)

= [x^3 +y^3 + 3x^2y + 3xy^2] + [y^3+z^3 + 3y^2z + 3yz^2] + [z^3+ x^3 + 3z^2x + 3zx^2] - 3[(xy+xz+y^2+yz) (x+z)]

= [x^3+x^3 +y^3+y^3+ z^3+ z^3] + 3x^2y + 3xy^2 + 3y^2z + 3yz^2 + 3z^2x + 3zx^2 - 3x^2y - 3xyz - 3x^2z - 3xz^2- 3y^2z - 3y^2x - 3xyz - 3yz^2

= (2x^3+ 2y^3+ 2z^3 - 6xyz)

= 2(x^3 +y^3 +z^3 - 2xyz)

Hope it helps uh!!