Question

If x=2-√3 y=√3-√7 and z=√7-√4 find the value of x³+y³+z³......plzz ans fast

Answer 1

check to add
$x + y + z$
if it becomes zero,then
${x}^{3} + {y}^{3} + {z}^{3} = 3xyz \\ 2 - \sqrt{3} + \sqrt{3} - \sqrt{7} + \sqrt{7} - \sqrt{4} \\ = 2 - \sqrt{4} \\ = 2 - 2 \\ = 0 \\ so \: condition \: is \: true \\ that \: means \:{x}^{3} + {y}^{3} + {z}^{3} = 3xyz \: \\ {x}^{3} + {y}^{3} + {z}^{3} = 3(2 - \sqrt{3} )( \sqrt{3} - \sqrt{7} )( \sqrt{7} - 2) \\ = 3(2 \sqrt{3} - 2 \sqrt{7} - 3 + \sqrt{21} )( \sqrt{7} - 2) \\ = 3(2 \sqrt{21} - 4 \sqrt{3} - 14 + 4 \sqrt{7} - 3 \sqrt{7} + 6 + \sqrt{21 \times 7} - 2 \sqrt{21} ) \\ = 3( - 4 \sqrt{3} - 14 + 6 + \sqrt{7} + 7 \sqrt{3} ) \\ = 3(3 \sqrt{3} + \sqrt{7} + 8) \\ {x}^{3} + {y}^{3} + {z}^{3} = 3(3 \sqrt{3} + \sqrt{7} + 8)$