Japan
If ^@ x = 1 + \sqrt{ 2 } + \sqrt{ 3 }, ^@ then ^@ x^4 - 4x^3 -4x^2 + 16x = ? ^@
Answer:
^@ 8 ^@
- Given: ^@ x = 1 + \sqrt{ 2 } + \sqrt{ 3 } ^@
According to the question, we need to find the value of ^@ x = 1 + \sqrt{ 2 } + \sqrt{ 3 } ^@ - ^@ \begin{align} & x = 1 + \sqrt{ 2 } + \sqrt{ 3 } \\
\implies & x - 1 = \sqrt{ 2 } + \sqrt{ 3 } \end{align} ^@
Taking the square on both sides
^@ \begin{align} & x^2 + 1 - 2x = 2 + 3 + 2 \times \sqrt{ 2 } \times \sqrt{ 3 } \space\space\space\space\space [ \text{ Using } (a + b)^2 = a^2 + b^2 + 2ab ] \\ \implies & x^2 - 2x - 4 = 2 \sqrt{ 6 } \\ \end{align} ^@ - Again, taking the square on both sides
^@ \begin{align} & \space x^4 + 4x^2 + 16 - 4x^3 + 16x - 8x^2 = 4 \times 6 \space\space\space\space\space [\because (a - b - c)^2 = a^2 + b^2 + c^2 -2ab + 2bc -2ca ] \\ & \implies x^4 - 4x^3 -4x^2 + 16x + 16 = 24 \\ & \implies x^4 - 4x^3 -4x^2 + 16x = 8 \\ \end{align} ^@
