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The square root of 27b^11

Can someone explain to me how to solve this?

  • Math -

    first, we factor the radicand (the term inside the radical sign or the squreroot):
    sqrt[ (3*3*3) (b*b^10) ]
    then we separate the perfect squares:
    sqrt[ 3*(3^2) b*((b^5)^2) ]
    and we take the squarerrot of the perfect squres. the squareroot of 3^2 is 3, while (b^5)^2 is b^5, then we can take them out of the sqrt, leaving only the 3 and b inside:
    3b^5 * sqrt(3b)

    hope this helps~ :)

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