The L.C.M. of $x^{3} - y^{3}, x - y$ and $x^{2} + x y + y^{2}$ will be

x^{2} - y^{2}

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

x - y

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

x^{3} - y^{3}

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

x (x + y)

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

Open in App

Solution

The correct option is C $x^{3} - y^{3}$
Since, $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$
$∴$ L.C.M. of $x^{3} - y^{3}, x - y$ and $x^{2} + x y + y^{2} = (x - y) (x^{2} + x y + y^{2}) = x^{3} - y^{3}$
Option C is correct.

Suggest Corrections

Similar questions

(x + y) + (x^{2} + x y + y^{2}) + (x^{3} + x^{2} y + x y^{2} + y^{3}) + . . . n

=

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{(x - y)^{2} + 3 x y}] + [\frac{(x + y)^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x 3 - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

∣ x ∣< 1

∣ y ∣< 1

(x + y) + (x^{2} + x y + y^{2}) + (x^{3} + x^{2} y + x y^{2} + y^{3}) + . . . . . \infty

Join BYJU'S Learning Program