Math, asked by vaishnavivaishu5323, 1 month ago

3. (a) If x+y+z = 0, show that x2 + y2 + z3 = 3xyz. (b) Show that (a - b)2 + (6 - c)+ (c-a) = 3 (a - b)(b-c)(c-a) ​

Answers

Answered by pushkardigraskar2005
3

Answer:

Here's your answer

Step-by-step explanation:

Given x + y + z = 0

⟹x + y = −z

Cubing on both sides

(x+y)^3 = (−z)^3

⟹x^3 + y^3 + 3x2y + 3xy2 = −z^3

⟹x^3 + y^3 + 3xy(x+y) = −z^3

⟹x^3 + y^3 + 3xy(−z) = −z^3

⟹x^3 + y^3 − 3xyz = −z^3

⟹x^3 + y^3 + z^3 = 3xyz

Hope you understand.

Answered by KB8
5

Given

x + y + z =0

⟹x + y = −z

Cubing on both sides

(x + y)³ = (−z)³

⟹x³ + y³  + 3x²y + 3xy² = −z³

⟹x³  + y³  + 3xy (x + y) = −z³  

⟹x³ + y³+ 3xy (−z) = −z³

⟹x³ + y³  − 3xyz = −z³

 ⟹x³ + y³  + z³  = 3xyz

∴ Hence Proved.

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Hope the answer helped you.

Thanks!!!

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