Question

3^5 sqrt (x+2)^3+3=27

Answer 1

where does the sqrt end?

Answer 2

sqrt (x+2)^3

Answer 3

\[3^5 \sqrt{(x+2)^3}+3=27~~~~~~~this?~~~~or,~~~~3\sqrt[5]{(x+2)^3}+3=27\]

Answer 4

the second one

Answer 5

I got disconnect, sorry.

Answer 6

its ok

Answer 7

@hartnn

Answer 8

Subtract 3 from both sides\[3\sqrt[5]{(x+2)^3}=24\]

Divide both sides by 3\[\sqrt[5]{(x+2)^3}=8\]

\[Let~~~~(x+2)=a~~~~~~~So,\]\[\sqrt[5]{a^3}=8\]

(This is not it, I'll post this so that you can process is while I am doing the rest.)

Answer 9

uh., let her try something on her own :P

Answer 10

Sorry

Answer 11

idk how to do it @hartnn !!!!!!!!

Answer 12

lol

Answer 13

thats why we are here,
tell us which step u don't get, which step are you stuck on, when u try to do it on your own

Answer 14

everything lol

Answer 15

Are you good so far?

Answer 16

from what your showing me yes

Answer 17

I got disconnected again (kt's probably the latex's fault.So
\[a^{3/5}=8~~~~~~~~~->~~~~~~~~~~~~~~~~a=8^{5/3}\]\[a=\sqrt[3]{8^5}~~~~~~~~->~~~~~~~a=\sqrt[3]{8^3 \times 8^2}=\sqrt[3]{(2^3)}~^5= 2^5 = 32\]