Q.If \(x, y, z\) are in continued proportion, then what is the value of \[ x^2y^2z^2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)? \] (a) \(x+y+z\) (b) \(x^2+y^2+z^2\) (c) \(x^3+y^3+z^3\) (d) None of the above
Answer: C
Let \(\frac{x}{y} = \frac{y}{z} = k\), where \(k \ne 0\) \(\therefore x = yk = zk^2,\quad y = zk\) \(\therefore x^2y^2z^2 \left(\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}\right)\) \(= z^2k^4 \cdot z^2k^2 \cdot z^2 \left(\frac{1}{z^3k^6} + \frac{1}{z^3k^3} + \frac{1}{z^3}\right)\) \(= z^6k^6 \left(\frac{1}{z^3k^6} + \frac{1}{z^3k^3} + \frac{1}{z^3}\right)\) \(= z^3 + z^3k^3 + z^3k^6\) \(= z^3 + (zk)^3 + (zk^2)^3\) \(= z^3 + y^3 + x^3\) \(= x^3 + y^3 + z^3\)
Similar Questions