If \((2x-6)^3+(x-8)^3+(x-22)^3=(6x-18)(x-8)(x-22)\), then find the value of \(x^2\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 81
The identity \(a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\) shows that \(a^3+b^3+c^3=3abc\) holds only when \(a+b+c=0\) or \(a=b=c\).
Take \(a=2x-6\), \(b=x-8\), \(c=x-22\)
Right side: \((6x-18)(x-8)(x-22)=3(2x-6)(x-8)(x-22)=3abc\)
\(b-c=(x-8)-(x-22)=14\neq0\), so \(a=b=c\) is impossible
Hence \(a+b+c=0\): \((2x-6)+(x-8)+(x-22)=4x-36=0\), so \(x=9\)
\(x^2=81\)