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Mathematics

Factorise :

(a2+1)b2b4a2(a^2 + 1) b^2 - b^4 - a^2

Factorisation

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Answer

(a2+1)b2b4a2=a2b2+b2b4a2=(a2b2a2)+(b2b4)=a2(b21)+b2(1b2)=a2(b21)b2(b21)=(b21)(a2b2)=(b1)(b+1)(ab)(a+b)(a^2 + 1) b^2 - b^4 - a^2\\[1em] = a^2b^2 + b^2 - b^4 - a^2\\[1em] = (a^2b^2 - a^2) + (b^2 - b^4)\\[1em] = a^2(b^2 - 1) + b^2(1 - b^2)\\[1em] = a^2(b^2 - 1) - b^2(b^2 - 1)\\[1em] = (b^2 - 1)(a^2 - b^2)\\[1em] = (b - 1)(b + 1)(a - b)(a + b)

Hence, (a2+1)b2b4a2=(b1)(b+1)(ab)(a+b)(a^2 + 1) b^2 - b^4 - a^2 = (b - 1)(b + 1)(a - b)(a + b).

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