Aspire Faculty ID #12099 · Topic: CUET 2025 · Just now
CUET 2025

If $x=(2+\sqrt{3})^{\frac{1}{3}}+(2+\sqrt{3})^{-\frac{1}{3}}$ and $x^{3}-3x+k=0$, then the value of k is:

Solution

Find k given \(x=(2+\sqrt{3})^{1/3}+(2+\sqrt{3})^{-1/3}\) and \(x^{3}-3x+k=0\)
  1. Let \(a=2+\sqrt{3}\Rightarrow a^{-1}=2-\sqrt{3}\). Define \(u^3=a,\;v^3=a^{-1}\) so that \(uv=\big(a\cdot a^{-1}\big)^{1/3}=1\) and \(x=u+v\).
  2. Use \((u+v)^3=u^3+v^3+3uv(u+v)\):
    \[ x^3=u^3+v^3+3uv(u+v)=(2+\sqrt{3})+(2-\sqrt{3})+3x=4+3x. \]
  3. Rearrange: \(\;x^3-3x-4=0\). Comparing with \(x^{3}-3x+k=0\) gives \(\;k=-4\).
Final answer: \(\boxed{k=-4}\)
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