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in Polynomials by (40.8k points)
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Factroise
`(i) 2a^7-128a`
`(ii) a^7+ab^6`
`(iii) 32a^3+108b^3`

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We have
`(i)( 2a^7-128a)`
` =2axx (a^6-64)=2a xx[(a^3)^2-8^2]`
`=2axx (a^3-8)xx(a^3+8) " " [because x^2-y^2=(x-y)(x+y)]`
`=2axx(a^3-2^3)xx(a^3+2^3)`
`2a xx[(a-2)xx (a^2+2a+4)]xx {(a+2)xx (a^2-2a +4)]`
`=2a (a-2) (a+2)(a^2+2a+4)(a^2-2a+4)`.
`therefore (2a^7-128a)=2a(a-2)(a+2)9a^2+2a+4)(a^2-2a+4)`
`(ii) (a^7+ab^6)`
`=axx(a^6xx b^6)=a xx {(a^2)^3+(b^2)^3}`
`=a xx (a^2+b^2)xx (a^4-a^2 b^2+b^4)" "[because x^3+y^3=(x+y)(x^2-xy+y^2)]`.
`therefore (a^7+ab^6)=a(a^2+b^2)(a^4-a^2 b^2+b^4)`.
`(iii) (32a^3+108b^3)`
`=4xx(8a^3+27b^3)`
`=4xx{(2a)^3+(3b)^3}`
`=4xx(2a+3b)xx (4a^2-6ab+9b^2) " " [because x^3+y^3=(x+y)(x^2-xy+y^2)]`
`therefore (32a^3+108b^3)=4(2a+3b)(4a^2-6ab+9b^2)`.

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