Question

If,x=1/3-^8 find the value of x^3-2x^2-7x+5

Answer 1

no I don't know

please I am sorry

Answer 2

$x = \frac{1}{3 - \sqrt{8} } = \frac{1}{3 - \sqrt{8} } \times \frac{(3 + \sqrt{8}) }{(3 + \sqrt{8} )} = \frac{(3 + \sqrt{8} }{(3)^{2} - ( \sqrt{8} )^{2} } \\ = \frac{(3 + \sqrt{8} }{9 - 8} = (3 + \sqrt{8} )$

$x = 3 + \sqrt{8} =>x - 3 = \sqrt{8} \\ =>(x - 3)^{2} = \sqrt{( {8})^{2} } = 8 \\ => {x}^{2} + 9 - 6x = 8=> {x}^{2} - 6x + 1 = 0$

${x}^{3} - {2x}^{2} - 7x + 5 = x( {x}^{2} - 6x + 1) + 4( {x}^{2} 6x + 1) + 16x + 1 \\ = x \times 0 \times + 4 \times 0 + 16(3 + \sqrt{8} ) + 1 \\ = 48 + 16 \sqrt{8} + 1 = 49 + 32 \sqrt{2}$