Given: nbsp; nbsp;x^2 8/x =x^3 8/x=1/x(x^3 8) [Taken 1/x common]=1/x(x^3 (2)^3) nbsp; =1/x(x 2)(x^2+2^2+2x) nbsp; nbsp;(using a^3 b^3=(a b)(a^2+b^2+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Factorisation Using Algebraic Identities

Factorise x2-...

Question

Factorise: $x^{2} - \frac{8}{x}$

$(x - 2) (x^{2} + 2 x + 4$ )

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

$\frac{1}{x}$ $(x - 2) (x^{2} + 2 x + 4$ )

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

$\frac{1}{x}$ $(x - 2) (x^{2} + 4$ )

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

$\frac{1}{x}$ $(x - 2) (2 x + 4$ )

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

Open in App

Solution

Given:
$x^{2} - \frac{8}{x}$

$= \frac{x^{3} - 8}{x}$
$= \frac{1}{x}$ $(x^{3} - 8$ ) [Taken $\frac{1}{x}$ common]
$= \frac{1}{x}$ $(x^{3} - (2)^{3}$ )
$= \frac{1}{x}$ $(x - 2) (x^{2} + 2^{2} + 2 x$ ) (using $a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b$ ))
$= \frac{1}{x}$ $(x - 2) (x^{2} + 4 + 2 x$ )
$∴ x^{2} - \frac{8}{x} = \frac{1}{x}$ $(x - 2) (x^{2} + 2 x + 4$ )
Hence, Option $A$ is correct.

Suggest Corrections

Factorise: x2−8x

Given: x2−8x =x3−8x=1x(x3−8) [Taken 1x common]=1x(x3−(2)3) =1x(x−2)(x2+22+2x) (using a3−b3=(a−b)(a2+b2+ab))=1x(x−2)(x2+4+2x)∴x2−8x=1x(x−2)(x2+2x+4)Hence, Option A is correct.

Factorise: $x^{2} - \frac{8}{x}$