it is either-
i) ³√(3-2√2)
ii) √(3-2√2)
iii) √(2-√2³√3+³√9) * √(3-2√2)
iv) (2-√2³√3 + ³√9) * √(3-2√2)
First of all, multiplying √(√2 + ³√3) by √(√2 + ³√3) yields (√2 + ³√3).
Next, we remove the square root by multiplying by (√2 - ³√3):
(√2 + ³√3) * (√2 - ³√3) = 2 + ³√9.
To remove the cube root, use difference of cubes x^3 - y^3 = (x - y)(x^2 + xy + y^2):
(2 + ³√9)(2^2 + 2 * ³√9 + (³√9)^2) = 2^3 - 9 = -1, which is rationalized.
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Overall rationalising factor:
√(√2 + ³√3)(√2 - ³√3)(4 + 2 * ³√9 + ³√81).
I hope this helps!
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Verified answer
First of all, multiplying √(√2 + ³√3) by √(√2 + ³√3) yields (√2 + ³√3).
Next, we remove the square root by multiplying by (√2 - ³√3):
(√2 + ³√3) * (√2 - ³√3) = 2 + ³√9.
To remove the cube root, use difference of cubes x^3 - y^3 = (x - y)(x^2 + xy + y^2):
(2 + ³√9)(2^2 + 2 * ³√9 + (³√9)^2) = 2^3 - 9 = -1, which is rationalized.
--------------
Overall rationalising factor:
√(√2 + ³√3)(√2 - ³√3)(4 + 2 * ³√9 + ³√81).
I hope this helps!