The factor of $x^{8} + x^{4} + 1$ are

(x^{4} + 1 - x^{2}), (x^{2} + 1 + x), (x^{2} + 1 - x)

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

(x^{4} + 1 - x^{2}), (x^{2} 11 + x), (x^{2} + 1 + x)

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

(x^{4} - 1 + x^{2}), (x^{2} - 1 + x), (x^{2} + 1 + x)

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

(x^{4} - 1 + x^{2}), (x^{2} + 1 - x), (x^{2} + 1 + x)

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

Open in App

Solution

The correct option is C $(x^{4} + 1 - x^{2}), (x^{2} + 1 + x), (x^{2} + 1 - x)$
$x^{8} + x^{4} + 1$

$= (x^{4})^{2} + (1)^{2} + x^{4}$
$= (x^{4} + 1)^{2} - 2 x^{4} + x^{4}$
$= (x^{4} + 1)^{2} - (x^{2})^{2}$
$= (x^{4} + 1 + x^{2}) (x^{4} + 1 - x^{2})$
$= [(x^{2})^{2} + (1)^{2} + x^{2}] [x^{4} - x^{2} + 1]$
$= [(x^{2} + 1)^{2} - 2 x^{2} + x^{2}] [x^{4} - x^{2} + 1]$
$= [(x^{2} + 1)^{2} - (x)^{2}] [x^{4} - x^{2} + 1]$
$= (x^{2} + 1 + x) (x^{2} + 1 - x) ((x^{4} - x^{2} + 1)$
$= (x^{2} + x + 1) (x^{2} - x + 1) (x^{4} - x^{2} + 1)$
Option A is correct.

Suggest Corrections

Similar questions

\frac{2 x}{1 + x^{2} + x^{4}} + \frac{1}{1 + x + x^{2}} - \frac{1}{1 - x + x^{2}} =

\frac{x}{1 - x^{2}} + \frac{x^{2}}{1 - x^{4}} + \frac{x^{4}}{1 - x^{8}} + . . . . .

\infty

x + (\frac{1}{x}) = 6

x^{2} + (\frac{1}{x^{2}})

x^{4} + (\frac{1}{x^{4}})

lim x \to 0 \frac{x (e^{(\sqrt{1 + x^{2} + x^{4}} - 1) / x} - 1)}{\sqrt{1 + x^{2} + x^{4}} - 1}

\frac{x}{1 - x^{2}} + \frac{x^{2}}{1 - x^{4}} + \frac{x^{4}}{1 - x^{8}} + . . . .

| x | < 1

Join BYJU'S Learning Program