(i) (2a + 36 + 4c)(4a2 + 9b2 + 16c2 – 6ab – 12bc – 12bc – 8ca)
We know that
(a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c x 3 abc
∴ (2a + 36 + 4c) (4a2 + 9b2 + 16c2 – 6 ab – 12 bc – 8 ca)
= (2a)3 + (3b)3 + (4c)3 – 3 x 2a x 36 x 4c
= 8a3 + 27b3 + 64c3 – 72 abc
(ii) (x – 2,y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3xz)
(a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc.
∴ (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz – 3xz)
= x3 + (-2y)3 + (3z)3 – 3 x (x) x (-2y) (3z)
= x3 – 8y3 + 27z3 + 18 xyz
