The value of $(1 + w^{2} + 2 w)^{3 n} - (1 + w^{2} + 2 w^{2})^{3 n}$ is equal to

Zero

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

w

No worries! We‘ve got your back. Try BYJU‘S free classes today!

w^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A Zero
$(1 + ω^{2} + 2 ω)^{3 n} - (1 + ω + 2 ω^{2})^{3 n}$

$∵ ω^{2} + ω + 1 = 0$
$\Rightarrow ω^{2} + 2 ω + 1 = ω$

$1 + ω + 2 ω^{2} = ω^{2}$
$\Rightarrow (ω)^{3 n} - (ω^{2})^{3 n} = (ω^{3})^{n} - (ω^{3})^{2 n} = 1 - 1 = 0$ .

Suggest Corrections

Similar questions

w

w^{2}

{(1 + 3 w + w^{2}) + (1 + 3 - w^{2})}^{4}

1, w, w^{2}

(1 - w + w^{2}) (1 + w - w^{2})

If 1, w, $w^{2}$ are three cube roots of unity, then $(1 - w + w^{2}) (1 + w - w^{2})$ is .............

1, w, w^{2}

{(1 - w + w^{2})}^{6} + {(1 - w^{2} + w)}^{6} = 128

If 1, w, $w^{2}$ are three cube roots of unity, then $(1 - w + w^{2}) (1 + w - w^{2})$ is .............

Join BYJU'S Learning Program