Question

(a) If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
(b) Show that (a - b)³ + (b - c)³ +(c - a)³ = 3 (a - b) (b - c) (c-a)​

Answer 1

Step-by-step explanation:

a)THERE IS A FORMULA i.e.

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

FROM THIS FORMULA IF (x+y+z)= 0 then the formula will be:-

x^3+y^3+z^3 -3xyz = 0

x^3+y^3+z^3=3xyz(SHIFTING -3xyz)(Proved)

b) We know that x^3+y^3+z^3=3xyz ( if x+y+z=0)

HERE,

x=a-b ; y=b-c ; z= c-a

Adding = x+y+z= a-b+b-c+c-a=0( every variable will be cancelled)

so the ans will be 3xyz= 3(a - b)(b-c)(c-a)

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