Question

factorise a^3+27b^3+8c^3-18abc

Answer 1

hey dear✨

answer is ........



(a)³+(3b)³+(2c)³-3(a)(3b)(2c)
using formula a³+b³+c³-3abc = (a+b+c)(a²+b²+ c²-ab-bc-ca)
(a+3b+2c)(a²+9b²+4c²-3ab-6bc-2ac)

hope this helps

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Here we have :-

a^3 + (3b)^3 + (2c)^3 - 3 × a × (3b) × (2c)

Identity :-

x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)

★Here we have :-

a = x , 3b = y and 2c = z

★Substitute the values we get :-

= (x + y + z)(x^2+ y^2 + z^2 - xy - ya - zx)

= (a + 3b + 2c)(a^2 + 9b^2 + 4c^2 - 3ab - 6bc - 2ca)