Q49 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics

If $1, \omega, \omega^2$ are the cube roots of unity and $(x+y)(x \omega+$ $\left.y \omega^2\right)\left(x \omega^2+y \omega\right)=f(x, y)$, then $f(2,3)=$

  • A
    16
  • B
    24
  • Correct answer
    C
    35
  • D
    45

Solution

$\begin{aligned} & \text { Since, } f(x, y)=(x+y)\left(x \omega+y \omega^2\right)\left(x \omega^2+y \omega\right) \\ & \Rightarrow f(2,3)=(2+3)\left(2 \omega+3 \omega^2\right)\left(2 \omega^2+3 \omega\right) \\ & =5\left(4 \omega^3+6 \omega^2+6 \omega^4+9 \omega^3\right) \\ & =5\left(4+6 \omega^2+6 \omega+9\right) \\ & =5\left(13+6\left(\omega^2+\omega\right)\right)=5(13+6 \times(-1)) \quad\left[\because \omega^3=1\right] \\ & =5 \times 7=35\end{aligned}$

Hence, the correct answer is option: 'C'.

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