If \(\left(x+\displaystyle\frac{1}{2}\right)^3+\left(3x+\frac{2}{3}\right)^3+\left(x-\frac{1}{6}\right)^3=\frac{1}{12}(9x+2)(2x+1)(6x-1)\), then find the value of \(x^2\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{25}\)
Take \(p=x+\frac{1}{2}\), \(q=3x+\frac{2}{3}\), \(r=x-\frac{1}{6}\). The right side turns out to be \(3pqr\), and \(p^3+q^3+r^3-3pqr=(p+q+r)(p^2+q^2+r^2-pq-qr-rp)\). So the equation holds when \(p+q+r=0\), or when the second factor is zero, which happens only if \(p=q=r\).
\(p=\displaystyle\frac{2x+1}{2}\), \(q=\displaystyle\frac{9x+2}{3}\), \(r=\displaystyle\frac{6x-1}{6}\)
\(3pqr=3\times\displaystyle\frac{(2x+1)(9x+2)(6x-1)}{36}=\frac{1}{12}(9x+2)(2x+1)(6x-1)\), the right side
\(p=r\) would need \(\displaystyle\frac{1}{2}=-\frac{1}{6}\), which is false, so \(p=q=r\) is not possible
Hence \(p+q+r=0\): \(x+\displaystyle\frac{1}{2}+3x+\frac{2}{3}+x-\frac{1}{6}=5x+1=0\)
\(x=-\displaystyle\frac{1}{5}\), so \(x^2=\displaystyle\frac{1}{25}\)
Why the other options are incorrect
Option (1) \(-\frac{1}{5}\) — Incorrect. This is the value of \(x\); the question asks for \(x^2\).
Option (3) \(-\frac{1}{4}\) — Incorrect. A square of a real number cannot be negative.
Option (4) \(-\frac{2}{3}\) — Incorrect. A square of a real number cannot be negative.