Let a+b=A, and 2(a-b)=B, so the given equation can be restated:
(a+b)³-8(a-b)³=A³-B³ ··· Eq.1 Use the standard polynomial form for A³-B³. We have:
A³-B³=(A-B)(A²+AB+B²) ··· Eq.2 Plug A=a+b, and B=2(a-b) into each term of LHS. We have:
A-B=(a+b)-2(a-b)=3b-a
A²=(a+b)²=a²+2ab+b²
AB=(a+b)x2(a-b)=2a²-2b²
B²=2²(a-b)²=4a²-8ab+4b² Thus, we have:
A²+AB+B²=7a²-6ab+3b² From Eq.1 and Eq.2, we have:
(a+b)³-8(a-b)³=(3b-a)(7a²-6ab+3b²) ··· Eq.3
CK: Plug a=2, b=1 (or any number) into Eq.3 LHS=(2+1)³-8(2-1)³=27-8=19 While,
RHS=(3(1)-2)(7(2)²-6(2)(1)+3(1)²)=(1)(28-12+3)=19 Thus, LHS=RHS CKD.
The answer: (a+b)³-8(a-b)³=(3b-a)(7a²-6ab+3b²)