Question

$x {?}^{2} + 1 \div x {?}^{2} = 7 \\find \: x {?}^{3} + 1 \div x {?}^{3} =what$

Answer 1

$\huge\boxed{\texttt{\fcolorbox{Red}{aqua}{Hello Friend}}}$


Here is your answer
$x {}^{3} + \frac{1}{x {}^{3} } = (x + \frac{1}{x})(x {}^{2} + \frac{1}{x {}^{2} } - x \times \frac{1}{x} ) \\ x {}^{3} + \frac{1}{x {}^{3} } = (x + \frac{1}{x} )(7 - 1) \\ x {}^{3} + \frac{1}{x {}^{3} } = (x + \frac{1}{x} ) \times 6 \\ now \: squaring \: both \: sides \\ (x {}^{3 } + \frac{1}{x {}^{3} } ) {}^{2} = (x + \frac{1}{x} ) {}^{2} \times 6 {}^{2} \\ (x {}^{3} + \frac{1}{x {}^{3} } ) {}^{2} = (x {}^{2} + \frac{1}{x {}^{2} } + 2 \times x \times \frac{1}{x} ) \times 36 \\ (x {}^{3} + \frac{1}{x {}^{3} } ) {}^{2} = (7 + 2) \times 36 \\ x {}^{3} + \frac{1}{x {}^{3} } = \sqrt{9 \times 36} \\ x {}^{3} + \frac{1}{x {}^{3} } = \sqrt{324} \\ x {}^{3} + \frac{1}{x {}^{3} } = 18$
Hope it helps you