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प्रश्न
If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, prove that (x + y + z)3 = 27xyz
उत्तर
`x^(1/3) + y^(1/3) + z^(1/3)` = 0
⇒ `(x^(1/3) + y^(1/3)) + z^(1/3)` = 0 cubing both sides, we get :
⇒ `(x^(1/3) + y^(1/3))^3 + z + 3 (x^(1/3) + y^(1/3)) z^(1/3) (x^(1/3) + y^(1/3) + z^(1/3))` = 0
⇒ `x + y+ 3 x^(1/3)y^(1/3) (x^(1/3) + y^(1/3)) + z + 0` = 0
⇒ `x + y+ 3 x^(1/3)y^(1/3)(-z^(1/3)) + z` = 0 ...(Using the given condition again)
⇒ x + y + z = `3 x^(1/3)y^(1/3)z^(1/3)`
⇒ (x + y + z)3 = 27xyz.
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