MyQuestionIcon

1

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

LCM

The LCM of ...

Question

The LCM of $6 (x^{2} + x y), 8 (x y - y^{2}), 12 (x^{2} - y^{2}) a n d 20 {(x + y)}^{2}$ is

A

120 x (x + y) (x - y)

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

B

120 x y (x + y) (x - y)

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

C

120 x y {(x + y)}^{2} (x - y)

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

D

120 x y (x + y) {(x - y)}^{2}

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

Open in App

Solution

The correct option is C $120 x y {(x + y)}^{2} (x - y)$
$6 (x^{2} + x y) = 6 x (x + y)$
$8 (x y - y^{2}) = 8 y (x - y)$
$12 (x^{2} - y^{2}) = 12 (x - y) (x + y)$
$20 {(x + y)}^{2} = 20 (x + y) (x + y)$
LCM of 6, 8, 12, 20 = 120
$∴$ Reqd LCM = $120 x y (x - y) {(x + y)}^{2}$

Suggest Corrections

thumbs-up

0

Similar questions

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

x^{y} = y^{x},

(\frac{x}{y})^{\frac{x}{y}}

is equal to:

b) If

\frac{x}{y} = \frac{6}{5}

\frac{x^{2} + y^{2}}{x^{2} - y^{2}}

Q. The factorisation of

2 (x + y)^{2} - 6 (x + y)

x^{y} = y^{x},

(\frac{x}{y})^{\frac{x}{y}}

is equal to:

b) If

\frac{x}{y} = \frac{6}{5}

\frac{x^{2} + y^{2}}{x^{2} - y^{2}}

(x + y) (2 x + y) (2 x - y) (x - y)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Special Factors and Multiples

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard VI Mathematics

Join BYJU'S Learning Program

CrossIcon