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- Thread starter hassam
- Start date
- Messages
- 133
- Reaction score
- 47
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- 38
x^(1/3) - 2x^(-1/3) = 1
It cn also b written as:
x^(1/3) - 2/x^(1/3) = 1
Now substituting x^(1/3) by y gives:
y - 2/y = 1
y(y - 2/y = 1) ---> y^2 - 2 = y
y^2 - y - 2 = 0
y^2 + y - 2y - 2 = 0
y(y + 1) - 2(y + 1) = 0
(y + 1)(y - 2) = 0
y + 1 = 0 and/or y - 2 = 0
y = -1, y = 2
y = x^(1/3) i.e.
[y = x^(1/3)]^3 ---> x = y^3
x = (-1)^3, x = (2)^3
x = -1, x = 8
It cn also b written as:
x^(1/3) - 2/x^(1/3) = 1
Now substituting x^(1/3) by y gives:
y - 2/y = 1
y(y - 2/y = 1) ---> y^2 - 2 = y
y^2 - y - 2 = 0
y^2 + y - 2y - 2 = 0
y(y + 1) - 2(y + 1) = 0
(y + 1)(y - 2) = 0
y + 1 = 0 and/or y - 2 = 0
y = -1, y = 2
y = x^(1/3) i.e.
[y = x^(1/3)]^3 ---> x = y^3
x = (-1)^3, x = (2)^3
x = -1, x = 8