Q. 4 E4.2( 14 Votes )

# Let’s find the L.

Answer :

factors of x^{4} + x^{2}y^{2} + y^{4}:

Add and subtract x^{2}y^{2} to get,

x^{4} + x^{2}y^{2} + y^{4} = x^{4} + x^{2}y^{2} + y^{4} + x^{2}y^{2} - x^{2}y^{2}

= x^{4} + 2x^{2}y^{2} + y^{4} - x^{2}y^{2}

As we know (a + b)^{2} = a^{2} + b^{2} + 2ab,

So,

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

= (x^{2} + y^{2})^{2} – (xy)^{2}

Apply the formula a^{2} – b^{2} = (a + b) (a-b) in (x^{2} + y^{2})^{2} – (xy)^{2}to get,

= (x^{2} + y^{2}-xy) (x^{2} + y^{2}+xy)

Now factorise x^{3}y + y^{4},

= y (x^{3} + y^{3})

Apply a^{3} + b^{3} = (a + b)(a^{2} + b^{2} – ab) in (x^{3} + y^{3})

= y (x+y)(x^{2} + y^{2} – xy)

Now factorise (x^{2}-xy)^{3},

= (x^{2}-xy)^{2}(x^{2}-xy)

Apply (a-b)^{2} = a^{2} + b^{2} – 2ab in (x^{2}-xy)^{2} to get,

= [(x^{2})^{2} + (xy)^{2} – 2x^{2}xy][x(x-y)]

= x^{2} [x^{2} + y^{2} – 2xy][x(x-y)]

As we know (a - b)^{2} = a^{2} + b^{2} - 2ab in ,

= x^{3} (x-y)^{2} (x-y)

= x^{3} (x-y)^{3}

So the LCM of x^{4} + x^{2}y^{2} + y^{4}, x^{3}y + y^{4}, (x^{2}-xy)^{3} is:

x^{3}y (x-y)^{3}(x+y)(x^{2} + y^{2} – xy) (x^{2} + y^{2}+xy)

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