Question

if x+y+z=0 , Show that X³+y³+z³=3xyz​

Answer 1

Answer:

[x + y + z = 0] and [x³ + y³ + z³ = 3xyz]

Proved!

Step-by-step explanation:

► To show that if [x + y + z = 0] then, [x³ + y³ + z³ = 3xyz]

• First thing is that there is an identity which says that:

If x + y + z = 0, then x³ + y³ + z³ = 3xyz

► To prove the above identity

Let x = '1'

Let y = '2'

Let z = '-3'

So, [x + y + z] ⇒ [1 + 2 + (-3)]

∴ x + y + z = 0  (as mentioned in the question)

Now,

[x³ + y³ + z³] ⇒ 1³ + 2³ + (-3)³ → 1 + 8 + (-27)

∴ x³ + y³ + z³ = -18

Now, 3xyz ⇒ 3 × 1 × 2 × -3

∴ 3xyz = -18

So now we have,

• x + y + z = 0
• x³ + y³ + z³ = -18
• 3xyz = -18

∴ [x + y + z = 0] and [x³ + y³ + z³ = 3xyz]

Answer 2

Step-by-step explanation:

using identity only if x plus y plus z is equal to 0, then x cube plus y cube plus z cube is equal to three xyz