How do you subtract #3\root[ 3] { 2} - 7\root [ 3] { 128} #?

Question

1719 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-08T01:01:21

#-25root(3)(2)#

Remember, that to add and subtract radicals you need to have the same radicals on both sides ex: #sqrt(2) - 5sqrt(2) = -4sqrt(2)#

Now to make them the same radicals we need to simplify #7root(3)(128)# by breaking down 128 in to it's prime factorization.

#128 = 2*64 = 2*2*32 = 2*2*2*16 = 2*2*2*2*8 = 2*2*2*2*2*4 = 2*2*2*2*2*2*2#

Next, we can rewrite #7root(3)(128)# with it's prime factorizations

#7root(3)(128) = 7root(3)(2^3*2^3*2)=2*2*7root(3)(2)=28root(3)(2)#

Finally, we can now subtract the two radicals

#3root(3)(2) - 28root(3)(2) = -25root(3)(2)#

Answer 2 · 2017-12-08T01:07:28

With radicals it helps if you can take powers from under the root sign.

Since these are third roots, we have to take out third powers:

#128=2^7=2xx2^6=2xx(2^2)^3#

So we can take out #2^2=4# and leave the other #2# under the root:

So: #root 3 128=4root 3 2#

The original problem then turns into:

#=3root3 2 - (7xx4)root 3 2#

And since the radicals are all the same, we can subtract:

#=(3-28)xx root 3 2=-21root 3 2#