Question

Factorise (x-a) cube + x-b whole cube + x-c whole cube - 3(x-a)(x-b)(x-c) =?

Answer 1

Let, (x-a) = p, (x-b) = q, (x-c) = r.

pq = (x-a)(x-b) = x²-ax-bx+ab

qr = (x-b)(x-c) = x²-bx-cx+bc

pr = (x-a)(x-c) = x²-ax-cx+ac

Now, we can write, p³+q³+r³-3pqr

= p³ + q³ + r³ - 3abc

= (p+q+r)(p²+q²+r²-pq-qr-pr)

= (3x-a-b-c) [x²-2ax+a²+x²-2bx+b²+x²-2cx+c²-(x²-ax-bx+ab)-(x²-bx-cx+bc)-(x²-ax-cx+ac)]

= (3x-a-b-c) (x²-2ax+a²+x²-2bx+b²+x²-2cx+c²-x²+ax+bx-ab-x²+bx+cx-bc-x²+ax+cx-ac)

= (3x-a-b-c) (a²+b²+c²-ab-ac-bc)