Question

if x=7+4√3, find xsq.+1/xsq. and xcube+1/xcube​

Answer 1

Answer:

$x { }^{2} + \frac{1}{x {}^{2} } + 2 = (x + \frac{1}{x} ) {}^{2} \\ x {}^{2} + \frac{1}{x {}^{2} } = (7 + 4 \sqrt{3 } + \frac{1}{7 + 4 \sqrt{3} } ) {}^{2} - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = ( \frac{49 + 48 + 56 \sqrt{3} + 1 }{7 + 4 \sqrt{3} } ) {}^{2} - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = ( \frac{98 + 56 \sqrt{3} }{7 + 4 \sqrt{3} } ) {}^{2} - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = (14) {}^{2} - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 194$

Half of the question is solved the remaining you should solve by using identity of

(x+1/x)^3 =x^3 +1/x^3 + 3(x)(1/x)(x+1/x)