Advertisements
Advertisements
Question
If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, prove that (x + y + z)3 = 27xyz
Advertisements
Solution
`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.
APPEARS IN
RELATED QUESTIONS
If 5x + 1 = 25x - 2, find the value of 3x - 3 × 23 - x.
Evaluate the following:
`(2^3 xx 3^5 xx 24^2)/(12^2 xx 18^3 xx 27)`
Evaluate the following:
`(8/27)^((-2)/3) - (1/3)^-2 - 7^0`
Evaluate the following:
`16^(3/4) + 2(1/2)^-1 xx 3^0`
Evaluate the following:
`sqrt(1/4) + (0.01)^(-1/2) - (27)^(2/3)`
Find the value of k in each of the following:
`(sqrt(9))^-7 xx (sqrt(3))^-5` = 3k
If `root(x)("a") = root(y)("b") = root(z)("c")` and abc = 1, prove that x + y + z = 0
If ax = by = cz and abc = 1, show that
`(1)/x + (1)/y + (1)/z` = 0.
Prove the following:
`("a"^"m"/"a"^"n")^("m"+"n"+1) ·("a"^"n"/"a"^1)^("n" + 1-"m").("a"^1/"a"^"m")^(1+"m"-"n")`
Prove the following:
(xa)b-c x (xb)c-a x (xc)a-b = 1
