Sqaure root of 5 (u-1)(u^5+u^4+u^3+u^2+u+1)

GIVEN:

[tex]5(u - 1)( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1)[/tex]

remember:

[tex] \sqrt{u} = {u}^{ \frac{1}{2} } [/tex]

And

[tex] {u}^{n} \times {u}^{m} = {u}^{n + m} [/tex]

SOLVE:

start by multiplying the factors:

[tex]5( ({u}^{6} + {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u ) - ( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1))[/tex]

simplify by combing like terms. Most terms subtract off, leaving:

[tex]5( {u}^{6} - 1)[/tex]

This can be factored, but it is not a perfect square, which is really what we need to take the square root.

[tex]5( {u}^{3} - 1)( {u}^{3} + 1) [/tex]

I'm not exactly sure what form they want the answer in...

so taking the square root:

[tex] \sqrt{5( {u}^{6} - 1) } = {(5( {u}^{6} - 1))}^{ \frac{1}{2} } [/tex]

so my best answer is:

[tex] {5}^{ \frac{1}{2} } \times {( {u}^{6} - 1)}^{ \frac{1}{2} } [/tex]
or the more factored form:

[tex] {5}^{ \frac{1}{2} } { ({u}^{3} - 1)}^{ \frac{1}{2} } { ({u}^{3} + 1 )}^{ \frac{1}{2} } [/tex]

I'm not sure how else to solve it. Taking the square root doesn't work out super well, so I left it in the most simple form I could.

sorry for not coming to a definitive answer!