Question

The L.C.M of 8x²y3 and - 2xy is​

Answer 1

Answer:

Factorizing x2y2 - x2 by taking the common factor 'x2' we get,

x2(y2 - 1)

Now by using the identity a2 - b2.

x2(y2 - 12)

= x2(y + 1) (y - 1)

Also, factorizing xy2 - 2xy - 3x by taking the common factor 'x' we get,

x(y2 - 2y - 3)

= x(y2 - 3y + y - 3)

= x[y(y - 3) + 1(y - 3)]

= x(y - 3) (y + 1)

Therefore, the L.C.M. of x2y2 - x2 and xy2 - 2xy - 3x is x2(y + 1) (y - 1) (y - 3).

3. Find the L.C.M. of x2 + xy, xz + yz and x2 + 2xy + y2.

Solution:

Factorizing x2 + xy by taking the common factor 'x', we get

x(x + y)

Factorizing xz + yz by taking the common factor 'z', we get

z(x + y)

Factorizing x2 + 2xy + y2 by using the identity (a + b)2, we get

= (x)2 + 2 (x) (y) + (y)2

= (x + y)2

= (x + y) (x + y)

Therefore, the L.C.M. of x2 + xy, xz + yz and x2 + 2xy + y2 is xz(x + y) (x + y)

Step-by-step explanation:

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Answer 2

Answer:

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