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Question
Factorise a3−8b3−64c3−24abc
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Solution
Here the given expression can be written as,
a3−8b3−64c3−24abc=(a)3+(−2b)3+(−4c)3−3(a)(−2b)(−4c)
Comparing with the given identity,
x3+y3+z3−3xyz≡(x+y+z)(x2+y2+z2−xy−yz−xz)
We get factor as
=(a−2b−4c)[(a)2−(−2b)2+(−4c)2−(a)(−2b)−(−2b)(−4c)−(−4c)(a)]
=(a−2b−4c)(a2+4b2+16c2+2ab−8bc+4ca)
a3−8b3−64c3−24abc=(a)3+(−2b)3+(−4c)3−3(a)(−2b)(−4c)
Comparing with the given identity,
x3+y3+z3−3xyz≡(x+y+z)(x2+y2+z2−xy−yz−xz)
We get factor as
=(a−2b−4c)[(a)2−(−2b)2+(−4c)2−(a)(−2b)−(−2b)(−4c)−(−4c)(a)]
=(a−2b−4c)(a2+4b2+16c2+2ab−8bc+4ca)
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