If \(x+\displaystyle\frac{1}{15x}=3\), then find the value of \(\displaystyle\frac{1}{2}\left(9x^3+\frac{1}{375x^3}\right)\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 118.8
Take \(p=x\) and \(q=\frac{1}{15x}\). Then \(p+q=3\) and \(pq=\frac{1}{15}\), and the bracket is a multiple of \(p^3+q^3=(p+q)^3-3pq(p+q)\).
\(q^3=\displaystyle\frac{1}{3375x^3}\) and \(\displaystyle\frac{1}{375x^3}=\frac{9}{3375x^3}=9q^3\)
So \(9x^3+\displaystyle\frac{1}{375x^3}=9p^3+9q^3=9\left(p^3+q^3\right)\)
\(p^3+q^3=3^3-3\times\displaystyle\frac{1}{15}\times3=27-0.6=26.4\)
\(9\times26.4=237.6\)
Required value \(=\displaystyle\frac{1}{2}\times237.6=118.8\)