Question

if x = a + b , y =aw + bw^2 and z = aw^2 + bw Find x^3 + y^3 + z^3 =?

Answer 1

x^3 + y^3 + z^3 = 3xyz ..... is this correct

Answer 2

where W = cube root of unity.

Answer 3

just want to check\[1+w+w^2=0\]\[w^2=-1-w\]right yahoo?

Answer 4

if \(w=1\) its trivial and \[x^3+y^3+z^3=3(a+b)^3\]

Answer 5

The answer should be \[3(a^3 + b^3)\]

Answer 6

so ur problem is for when \(w\neq1\)

Answer 7

but....there is no condition in the question)

Answer 8

\[y^3+z^3=w^3[(a+bw)^3+(aw+b)^3]=(a+bw)^3+(aw+b)^3\]can u simplify this?

Answer 9

Lol....Do u knw Such a Formula:

x^3 + y^3 + z^3 = 3xyz

Answer 10

no its not true

Answer 11

Why.....

Answer 12

\[x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)\]

Answer 13

Yup...nw i got it...thxx

Answer 14

yahoo this time i really want u to go into it and do that simplifiction plz

Answer 15

yes....i am Doing it....

Answer 16

\[a^3 + 3a^2bw + 3ab^2w^2 + b^3 w^3 + a^3w^3 + 3a^2w^2 + 3awb^2 +b^3\]

Answer 17

\[w^3 = 1\]

Answer 18

@mukushla

Answer 19

ok replace \(w^3\) with 1 and \(w^2\) with \(-1-w\)

Answer 20

simplify and tell me what u get

Answer 21

\[2a^3 + 2b^3 +3ab^2w - 3ab^2 - 3a^2 - 3a^2w\]

Answer 22

u most get this\[a^3 + 3a^2bw + 3ab^2w^2 + b^3 w^3 + a^3w^3 + 3a^2w^2b + 3awb^2 +b^3\\=2(a^3+b^3)-3a^2b-3ab^2\]and finally\[(a+b)^3+2(a^3+b^3)-3a^2b-3ab^2=3(a^3+b^3)\]

Answer 23

yes.....i would have made a mistake.....thxx..


\[\boldsymbol{\mathfrak{ MUKUSHLA}}\]

Answer 24

\[\large \color\green{\text{Welcome}}\]

Answer 25

i syudied..latex today.....

Answer 26

Which website...u use for it...link..plzz

Answer 27

http://www.wolframalpha.com/input/?i=simplify+%28a+%2B+b%29^3+%2B+%28b+E^%28%282+I+ \[Pi]%29%2F3%29+%2B+++++a+E^%28%284+I+\[Pi]%29%2F3%29%29^3+%2B+%28a+E^%28%282+I+\[Pi]%29%2F3%29+%2B+++++b+E^%28%284+I+\[Pi]%29%2F3%29%29^3

Answer 28

\[
(a+b)^3+\left(e^{\frac{2 i \pi }{3}} a+e^{-\frac{1}{3}
(2 i \pi )} b\right)^3+\left(e^{-\frac{1}{3} (2 i
\pi )} a+e^{\frac{2 i \pi }{3}} b\right)^3=\\ 3 a^3+3
e^{\frac{2 i \pi }{3}} a^2 b+3 e^{-\frac{1}{3} (2 i
\pi )} a^2 b+3 a^2 b+3 e^{\frac{2 i \pi }{3}} a
b^2+3 e^{-\frac{1}{3} (2 i \pi )} a b^2+3 a b^2+3
b^3=\\ 3 a^3+3 e^{\frac{2 i \pi }{3}} a^2 b+3
e^{-\frac{1}{3} (2 i \pi )} a^2 b+3 a^2 b+3
e^{\frac{2 i \pi }{3}} a b^2+3 e^{-\frac{1}{3} (2 i
\pi )} a b^2+3 a b^2+3 b^3=\\ 3 a^3+3 a^2 b+6 a^2 b
\cos \left(\frac{2 \pi }{3}\right)+3 a b^2+6 a b^2
\cos \left(\frac{2 \pi }{3}\right)+3 b^3= \\ 3
a^3+\frac{6}{2} (-1) a^2 b+3 a^2 b+\frac{6}{2} (-1)
a b^2+3 a b^2+3 b^3=3 a^3+3 b^3


\]