Question

Factorise 8a6+5a3+1
By using a3+b3+c3-3abc

Answer 1

Answer with explanation:

$8a^6+5a^3+1\\\\=(2a^2)^3+(\sqrt5}a)^3+1^3\\\\=(2a^2+\sqrt5}a+1)[(2a^2)^2+(\sqrt5}a)^2+1^2-2a^2\times\sqrt{5}a-\sqrt{5}a\times1+1\times2a^2]+3\times(2a^2)\times(\sqrt{5}a)\times 1\\\\=(2a^2+\sqrt5}a+1)[(4a^4+5a^2+1^2-2\sqrt{5}a^3-\sqrt{5}a+2a^2]+6\sqrt{5}a^3$

The Identity,that we have used here, is,

$a^3+b^3+c^3-3 a b c=(a+b+c)(a^2+b^2+c^2-ab-b c-ca)$