Find the value of (a - 2x )3 + ( b - 2x )3 + ( c - 2x )3 - 3(a - 2x)(b - 2x)(c - 2x), given that a + b + c = 6x. A3 B0 C1 D2 Show answer & explanationCorrect answer: BExplanationThe correct answer is option 2 i.e. 0. (a−2x)3+(b−2x)3+(c−2x)3−3(a−2x)(b−2x)(c−2x)(a - 2x)^3 + (b - 2x)^3 + (c - 2x)^3 - 3(a - 2x)(b - 2x)(c - 2x)(a−2x)3+(b−2x)3+(c−2x)3−3(a−2x)(b−2x)(c−2x) p=a−2x,q=b−2x,r=c−2xp = a - 2x,\quad q = b - 2x,\quad r = c - 2xp=a−2x,q=b−2x,r=c−2x We know : p3+q3+r3−3pqr=0if p+q+r=0p^3 + q^3 + r^3 - 3pqr = 0 \quad \text{if } p + q + r = 0p3+q3+r3−3pqr=0if p+q+r=0 Now, p + q + r = (a + b + c) - 6x = 6x - 6x = 0 ⇒(a−2x)3+(b−2x)3+(c−2x)3−3(a−2x)(b−2x)(c−2x)=0\Rightarrow (a - 2x)^3 + (b - 2x)^3 + (c - 2x)^3 - 3(a - 2x)(b - 2x)(c - 2x) = 0⇒(a−2x)3+(b−2x)3+(c−2x)3−3(a−2x)(b−2x)(c−2x)=0