SOLUTION: ^6√27a^2b^5 x ^6√6a^3b^6

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Question 371723: ^6√27a^2b^5 x ^6√6a^3b^6
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
root%286%2C+27a%5E2b%5E5%29%2Aroot%286%2C+6a%5E3b%5E6%29
To multiply these we use a property of all radicals: root%28a%2C+p%29%2Aroot%28a%2C+q%29+=+root%28a%2C+p%2Aq%29:
root%286%2C+27a%5E2b%5E5%2A6a%5E3b%5E6%29
which simplifies to:
root%286%2C+162a%5E5b%5E11%29
Next we try to simplify the radical. This is done by finding factors that are powers of 6 (since this is a 6th root). There is one power of 6:
root%286%2C+b%5E6%2A162a%5E5b%5E5%29
Next we use the property above (this time from right to left) to separate the power of 6:
root%286%2C+b%5E6%29%2Aroot%286%2C+162a%5E5b%5E5%29
which simplifies to:
b%2Aroot%286%2C++162a%5E5b%5E5%29

Important: Radical expressions (without a negative sign in front) must not be negative. So our answer should not be negative. Normally one would use absolute value to ensure a non-negative result. However, since the original problem had root%286%2C+27a%5E2b%5E5%29 and since the radicand of a even-numbered root, like 6th roots, must be positive, we know that b could never be negative. So we do not need to be concerned with using absolute value to ensure a non-negative result. If we had not known that b was non-negative, we would have to use
abs%28b%29%2Aroot%286%2C++162a%5E5b%5E5%29
to ensure the final expression is non-negative.