SOLUTION: Good Evening Tutor, I have two questions I was wondering if you could tell me if I got them correct? 1.) Evaluate if possible: cubed(3) sqrt of -8/27 I got - 1/6 is th

Algebra ->  Square-cubic-other-roots -> SOLUTION: Good Evening Tutor, I have two questions I was wondering if you could tell me if I got them correct? 1.) Evaluate if possible: cubed(3) sqrt of -8/27 I got - 1/6 is th      Log On


   

Question 100682: Good Evening Tutor,
I have two questions I was wondering if you could tell me if I got them correct?
1.) Evaluate if possible: cubed(3) sqrt of -8/27 I got - 1/6 is this correct?
2.) simplify by combining like terms. sqrt of 63 - 2 sqrt of 28 + 5 sqrt of 7.
I got 7 sqrt of 7 I think this is wrong but I am not sure where I am getting messed up I keep getting the same answer.
Any help is appreciated. Thank you.
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
cubed sqrt of -8/27
%28-8%2F27%29%5E3=-512%2F19683 we cube the number
sqrt%28-512%2F19683%29=16sqrt%28-2%29%2F81sqrt%283%29 we get the square root
%2816sqrt%28-2%29%2F81sqrt%283%29%29%28sqrt%283%29%2Fsqrt%283%29+%29so a square root wont be in the denominator
=16sqrt%28-6%29%2F343=%2816sqrt%286%29%2Ai%29%2F343 the imaginary number i=sqrt%28-1%29
.
sqrt(63)-2 sqrt(28)+5 sqrt of (7)
=sqrt(9*7)-2sqrt(7*4)+5sqrt(7)
=3sqrt7-4sqrt(7)+5sqrt(7)
=4sqrt(7)
Ed