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hello

\[ \huge x=2^{\frac{ 1 }{ 3 }} + 4^{\frac{ 1 }{ 3 }}\]
Show that
\[\huge x ^{3} = 6(1+x)\]

post your working out, ill look at it and spot the error.

i worked out the Right hand side is equal to 46

have you got a scanner?

no...

damn, have you got any way of reproducing the page of working you did so I can look at it?

yea ok...
OH YEA the right hand side isnt equal to 46

|dw:1359973881754:dw|

now i dont think you can do anything with the RHS

|dw:1359973981397:dw|

im pretty sure there is a shortcut to do this question?

do you get x^3 = 23.08393261?

im pretty sure we need to show it using indices

lol, i dont even know those. As an answer i get \[\pm 2.847322102\]

well i now completely lost with this quesiton :P

we are meant to prove that
\[x ^{3} = 6(1+x)\]
not find x

we not meant to find x but prove

its basic substitution

\[\huge =6(2^{\frac { 2 }{ 3 }}-2+4^{\frac{ 2 }{ 3 }})\]

yes

yea but the original question need to be
|dw:1359975317046:dw|

\[\huge x^2=2^{\frac{ 2 }{ 3 }}+2+4^{\frac{ 2 }{ 3 }}\]
our expresion
\[\huge x^2-4\]

you lost me there :P

oh i found an easier way

\[\huge (2^{\frac{ 1 }{ 3 }} + 4^{\frac{ 1 }{ 3 }})^{3}\]
That is the question...

a + b)3 = a3 + 3a2b + 3ab2 + b3

ha i think your right

ok i worked it out by myself but thanks so much for your help jonask