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
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
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.
.
.
Hope the answer helped you.
Thanks!!!
Similar questions