If \((3x-4)^3+(x-6)^3+(x-10)^3=(9x-12)(x-6)(x-10)\), then find the value of \(x^2\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 16
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 \(a=3x-4\), \(b=x-6\), \(c=x-10\)
Right side: \((9x-12)(x-6)(x-10)=3(3x-4)(x-6)(x-10)=3abc\)
\(b-c=(x-6)-(x-10)=4\neq0\), so \(a=b=c\) is impossible
Hence \(a+b+c=0\): \((3x-4)+(x-6)+(x-10)=5x-20=0\), so \(x=4\)
\(x^2=16\)