Question

Help please!!! will fan and medal if you help guide me in the right direction. Question below.

Answer 1

What is the solution to the equation?\[3\sqrt[5]{(x+2)^{3}}+3=27\]

Answer 2

@mathmale @math_man21

Answer 3

What would I start with?

Answer 4

for once i cant help

Answer 5

Do you know someone who can?

Answer 6

I would try working out the stuff inside the square root

Answer 7

*fifth root

Answer 8

how do i do that exactly?

Answer 9

I think (you might need to FOIL but I'm not sure you can do that with a set of 3). But otherwise you can try x^3 and 2^3

Answer 10

can i help.

Answer 11

ya anyone can i just dont get this crap

Answer 12

@pooja195

Answer 13

should i start with the +3 thingy

Answer 14

ok, \[3\sqrt[5]({x+2})^{3}=27\]
\[3\sqrt[5]{(x+2)^{3}}=27-3\]

Answer 15

okay i got that part. but i dont get the other part

Answer 16

\[\sqrt[5]{(x+2)^{3}}=24/3\]
\[[(x+2)^{3}]^{1/5}=8\]

Answer 17

how did you get the 1/5

Answer 18

\[(x+2)^{3}=8^{5}\]
\[(x+2)=2^{3\times5/3}\]

Answer 19

sqart5 means 1/5

Answer 20

ooooooo okay

Answer 21

\[(x+2)=2^{5}\]
\[x+2=32\]
\[x=32-2\]\[x=30\]

Answer 22

thank you sooooooooooooo much you are a life saver!!!!!!!!!!