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If `(9sqrt3+11sqrt2)^[1/3]=sqrta+sqrtb`,where a and b are rational with a > b.Then

Answer» `(9sqrt3+11sqrt2)^(1/3) =sqrta+sqrtb`
Taking cube both sides,
`(9sqrt3+11sqrt2) = (sqrta+sqrtb)^3``(9sqrt3+11sqrt2) = (sqrta)^3+(sqrtb)^3 + 3(sqrta)^2b+3a(sqrtb)^2`
`(9sqrt3+11sqrt2) = asqrta+bsqrtb+3asqrtb+3sqrtab`
`(9sqrt3+11sqrt2) = (a+3b)sqrta+(b+3a)sqrtb`
If we put, `a = 3 and b = 2`, it satisfies the above equation.
So, option `B` is the correct option.


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