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Question

Find the L.C.M. and H.C.F. of x4+2x3+3x2+2x+1,x31,x3+x2+x, and x2y+yx+y.

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Solution

x4+2x3+3x2+2x+1=(x2+x+1)(x2+x+1)
x31=(x1)(x2+x+1)
x3+x2+x=x(x2+x+1)
x2y+yx+y=y(x2+x+1)
L.C.M. =(x2+x+1)2(x1)xy
and, H.C.F. =x2+x+1.

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