Question

If x+y+z=0, xyz=1 then prove that (1-x³)‐¹+(1-y³)‐¹+(1-z³)‐¹

Answer 1

Given :-

• x + y + z = 0
• xyz = 1 .

To Find :-

• (1-x³)‐¹+(1-y³)‐¹+(1-z³)‐¹ ?

Formula used :-

• 1+ω+ω² = 0
• ω³ⁿ= 1 ( n = natural Number).
• ω = Omega .
• (a)‐¹ = 1/a .
• [(a)^b]^c = (a)^(b*c)
• 1/0 = ∞

Solution :-

→ x + y + z = 0 = 1+ω+ω²

Comparing we get :-

→ x = 1

→ y = ω

→ z = ω²

Putting These value we get :-

→ (1-x³)‐¹+(1-y³)‐¹+(1-z³)‐¹

→ 1/(1 - x³) + 1/(1 - y³) + 1/(1 - z³)

→ 1/( 1 - 1) + 1/(1 - ω³) + 1/{1 - (ω²)³}

→ 1/( 1 - 1) + 1/(1 - 1) + 1/{1 - (ω^6)}

→ 1/( 1 - 1) + 1/(1 - 1) + 1/(1 - 1)

→ 1/0 + 1/0 + 1/0

→ ∞

Hence, Required Answer is ∞ (infinity).