Question

prove that (1+w)^3-(1-w^2)^3=0?​

Answer 1

Step-by-step explanation:

Question is wrong it, should be

(1+w)^3-(1+W^2)^3=0

Solution:-

L.H.S

= (1+W)^3 -(1+W^2)^3

= (-w^2)^3 - (-w-w^2+w^2)^3

= (-w^6) - (-w^3)

=( -1) -(-1)

= -1 +1

= 0

= R.H.S

Answer 2

$(1+\omega)^3-(1+\omega^2)^3=0$, proved.

Step-by-step explanation:

To prove that, $(1+\omega)^3-(1+\omega^2)^3=0$

L.H.S . = $(1+\omega)^3-(1+\omega^2)^3$

We know that,

$1+\omega+\omega^2=0$

⇒ $1+\omega=-\omega^2$

⇒ $1+\omega^2=-\omega$

= $(-\omega^2)^3-(-\omega)^3$

= $-\omega^6+\omega^6$

= $-(\omega^3)^2+(\omega^3)^2$

= - 1 + 1 [ ∵ $\omega^3=1$]

= 0

=R.H.S., proved.