TeemoTheTerific 2 years ago hi Can someone help me with a question? Help much appreciated

1. Razzputin

hello

2. TeemoTheTerific

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

3. TeemoTheTerific

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)

4. Razzputin

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

5. TeemoTheTerific

i worked out the Right hand side is equal to 46

6. Razzputin

have you got a scanner?

7. TeemoTheTerific

no...

8. Razzputin

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

9. TeemoTheTerific

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

10. TeemoTheTerific

|dw:1359973881754:dw|

11. TeemoTheTerific

now i dont think you can do anything with the RHS

12. TeemoTheTerific

|dw:1359973981397:dw|

13. TeemoTheTerific

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

14. Razzputin

do you get x^3 = 23.08393261?

15. TeemoTheTerific

im pretty sure we need to show it using indices

16. Razzputin

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

17. TeemoTheTerific

well i now completely lost with this quesiton :P

18. Razzputin

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$

19. TeemoTheTerific

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

$x^3+y^3=(x+y)(x^2-xy+y^2)$ $\huge x^3=(2+4)(2^{\frac { 2 }{ 3 }}-8^{\frac{ 1 }{ 3 }}+4^{\frac{ 2 }{ 3 }})$

21. Razzputin

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

23. Razzputin

its basic substitution

24. TeemoTheTerific

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

yes

26. TeemoTheTerific

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 $\huge x^2-4$

28. TeemoTheTerific

you lost me there :P

oh i found an easier way

30. Kuoministers

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

31. Kuoministers

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

32. TeemoTheTerific

lets factor 2^1/3$\huge 2^{\frac{ 1 }{ 3 }}(1+2^{\frac{1}{3}})$ $\huge 2^{\frac{ 1 }{ 3 }3}(1+2^{\frac{ 1 }{ 3 }})^3$ $\huge 2(1+3(1)2^{1/3}+3(2^{2/3})+2)$ $\huge 2(3)(1+2^{1/3}+2^{2/3})$ $\huge 6(1+x)$

34. TeemoTheTerific

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