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Question

Factorise x6+5x3+8

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Solution

Given x6+5x3+8

Adding and subtracting x3

=x6+5x3+8+x3x3

=x6+6x3+8x3

=(x2)3+(2)3+(x)33(x2)(2)(x)

According to the identity
[x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)]

=(x2+2x)[(x2)2+(2)2+(x)2(x2)(2)(2)(x)(x2)(x)]

=(x2x+2)(x4+x2+42x2+2x+x3)

=(x2x+2)(x4+x3x2+2x+4)

Thus, factors of x6+5x3+8 are (x2x+2)(x4+x3x2+2x+4)

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