If \(x+\displaystyle\frac{1}{4x}=2\), then find the value of \(8x^3+\displaystyle\frac{1}{8x^3}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 52
Take \(p=x\) and \(q=\frac{1}{4x}\). Then \(p+q=2\) and \(pq=\frac{1}{4}\), and the expression is a multiple of \(p^3+q^3=(p+q)^3-3pq(p+q)\).
\(q^3=\displaystyle\frac{1}{64x^3}\) and \(\displaystyle\frac{1}{8x^3}=\frac{8}{64x^3}=8q^3\)
So \(8x^3+\displaystyle\frac{1}{8x^3}=8p^3+8q^3=8\left(p^3+q^3\right)\)
\(p^3+q^3=2^3-3\times\displaystyle\frac{1}{4}\times2=8-1.5=6.5\)
\(8\times6.5=52\)