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Mathematics
OpenStudy (anonymous):

Please help with this question. Please show every step of the solution. MEDALS WILL BE APLENTY!!!

Parth (parthkohli):

Lol we don't need medals :P

OpenStudy (lgbasallote):

you do know you cant vouch medals will be aplenty

OpenStudy (lgbasallote):

therefore shall we go away now?

Parth (parthkohli):

Bye!

OpenStudy (anonymous):

really soz ill post my question in a sec

OpenStudy (anonymous):

here it is

OpenStudy (anonymous):

thanks guys in advance

Parth (parthkohli):

Find the cube of both sides =.=

OpenStudy (lgbasallote):

ugh you couldnt have written that =_= :p lol anyway substitute x

OpenStudy (lgbasallote):

that should be your first step

OpenStudy (anonymous):

just take cube of both sides and factor 6 out lol

OpenStudy (anonymous):

yeh i have done all that but i can't seem to get the answer

Parth (parthkohli):

\(\Large \color{Black}{\Rightarrow (a + b)^3 = a^3 + b^3 + 3ab^2 + 3a^2b }\)

OpenStudy (anonymous):

It's not equal to 4! If you multiply it will equal 4

OpenStudy (anonymous):

Yes use the formula @ParthKohli gave you.

OpenStudy (lgbasallote):

substute x is still the first step...something is wrong with my algebra today @_@ lol

Parth (parthkohli):

@lgbasallote I can teach you algebra xD

OpenStudy (anonymous):

oh ok guys yep i got it. i love this website !!!XDXD

Parth (parthkohli):

Lol you should be loving the site

OpenStudy (lgbasallote):

i just answered 10 algebra questions wrong o.O what is wrong with me =_= lol

OpenStudy (anonymous):

\[\large (\sqrt[3]{2}+\sqrt[3]{4})^3=\sqrt[3]{2^3}+3*\sqrt[3]{2^2}*\sqrt[3]4+3*\sqrt[3]2*\sqrt[3]{4^2}+\sqrt[3]{4^3}\]\[\large=2+3*\sqrt[3]{16}+3*\sqrt[3]{16}+4\]\[\large=6+6*\sqrt[3]{16}=6(1+\sqrt[3]{16})\]That's the closest I can get.

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