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Question

x2(x2+a2)(x2+b2)=k[a2x2+a2b2x2+b2]k=

A
1
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B
1a2+b2
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C
1a2b2
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D
1b2a2
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Solution

The correct option is D 1a2b2
Writing the LHS in partial fractions
By substituting x2=z
z(z+a2)(z+b2)=Az+a2+Bz+b2
We get the values of A and B
By comparing co-efficients we get
A=a2a2b2 and B=b2a2b2
x2(x2+a2)(x2+b2)=a2(a2b2)(x2+a2)b2(a2b2)(x2+b2)
=1a2b2[a2x2+a2b2x2+b2]

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