Answer :

(a3)2 + 3a3b3-40(b3)2

To resolve it into factors we first resolve 40


40=2×2×2×5


Given, x2 + (m + n)x + mn=(x + m)(x + n) … (2)


Comparing 1 and 2


m + n =3


mn =-40


(1) can be written as


(a3)2 + 8a3b3-5a3b3-40(b3)2


= (a3 + 8b3)( a3-5b3)


The resolved factors are (a3 + 8b3)( a3-5b3)


Apply the formula a3 + b3 = (a+b)(a2 + b2 – ab) in (a3 + 8b3)


Now (a3 + 8b3) = (a3 + (2b)3)


= (a2 + 4b2 – 2ab)( a3-5b3)


The resolved factors are (a2 + 4b2 – 2ab)( a3-5b3).


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