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hi Can someone help me with a question? Help much appreciated

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\[ \huge x=2^{\frac{ 1 }{ 3 }} + 4^{\frac{ 1 }{ 3 }}\] Show that \[\huge x ^{3} = 6(1+x)\]
can you please show me all woking out as i tried it and it took up like half a page :P(prob wrong working out)

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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?
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
now i dont think you can do anything with the RHS
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
you were going the right way.\[x^{3} = 6(1+(2^\frac{ 1 }{ 3 }+4^\frac{ 1 }{ 3 }))\] \[x^{3} = 6(1+(\sqrt[3]{2}+\sqrt[3]{4}))\]\[x^{3} = 6(1+(2.847322102))\]\[x^{3} = 23.08393261\]\[x =\sqrt[3]{23.08393261}\]\[x = \pm2.84732210\]
we are meant to prove that \[x ^{3} = 6(1+x)\] not find x
\[x^3+y^3=(x+y)(x^2-xy+y^2)\] \[\huge x^3=(2+4)(2^{\frac { 2 }{ 3 }}-8^{\frac{ 1 }{ 3 }}+4^{\frac{ 2 }{ 3 }})\]
oh blast. lol well then if x^3 = 23.08393261 then substitute given in for x therefore \[(\sqrt[3]{2}+\sqrt[3]{4})^{3}\] what does it equal?
we not meant to find x but prove
its basic substitution
\[\huge =6(2^{\frac { 2 }{ 3 }}-2+4^{\frac{ 2 }{ 3 }})\]
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
lets factor 2^1/3\[\huge 2^{\frac{ 1 }{ 3 }}(1+2^{\frac{1}{3}})\] \[\huge 2^{\frac{ 1 }{ 3 }3}(1+2^{\frac{ 1 }{ 3 }})^3\] \[\huge 2(1+3(1)2^{1/3}+3(2^{2/3})+2)\] \[\huge 2(3)(1+2^{1/3}+2^{2/3})\] \[\huge 6(1+x)\]
ok i worked it out by myself but thanks so much for your help jonask

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