32a³+108b³=4x8a³+4x27b³=4(2a)³+4(3b)³
Let 2a=A, and 3b=B The given equation can be restated as follows:
32a³+108b³=4(A³+B³) ··· Eq.1 Use the standard polynomial form for A³+B³. We have:
A³+B³=(A+B)(A²-AB+B³) ··· Eq.2 Plug A=2a, B=3b into RHS of Eq.2. We have:
A³+B³=(2a+3b)(4a²-6ab+9b²) ··· Eq.3 From Eq.1 and Eq.3, we have:
32a³+108b³=4(2a+3b)(4a²-6ab+9b²) ··· Eq.4
CK: Plug a=1, b=2 (or any number) into Eq.4 LHS=32(1)³+108(2)³=32+864=896 While,
RHS=4(2(1)+3(2))(4(1)²-6(1)(2)+9(2)²)=4x8x28=896 Thus, LHS=RHS CKD.
The answer: 32a³+108b³=4(2a+3b)(4a²-6ab+9b²)