Question

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Answer 1

Answer:

6,18

Step-by-step explanation:

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Answer 2

$Given \: \: Question \: \: Is \: \: \: \\ \\ x + \frac{1}{x} = 3 \\ \\ 01 \\ \\ find \: \: x {}^{2} + \frac{1}{x {}^{2} } \\ \\ Answers \: \: \: \\ \\ x + \frac{1}{x} = 3 \\ \\ squaring \: \: on \: \: both \: \: sides \: \: we \: \: have \\ \\ (x + \frac{1}{x} ) {}^{2} = (3) {}^{2} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2x \frac{1}{x} = 9 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 9 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 9 - 2 \\ \\ x { }^{2} + \frac{1}{x {}^{2} } = 7 \\ \\ Therefore \: \: \: \: \: \: \: x {}^{2} + \frac{1}{x {}^{2} } = 7 \\ \\ \\ 02 \\ \\ find \: \: \: \: \: \: \: \: \: x {}^{3} + \frac{1}{x {}^{3} } \\ \\ Answer \: \\ \\ x + \frac{1}{x} = 3 \\ \\ cubing \: \: \: on \: \: both \: \: sides \: \: we \: \: have \\ \\ (x + \frac{1}{x} ) {}^{3} = (3) {}^{3} \\ \\ \\ x {}^{3} + \frac{1}{x {}^{3} } + 3(x + \frac{1}{x} ) = 27 \\ becoz \: \: \: (a + b) {}^{3} = a {}^{3} + b {}^{3} + 3ab(a + b) \\ \\ \\ x {}^{3} + \frac{1}{x {}^{3} } + 3(3) = 27 \\ \\ \\ x {}^{3} + \frac{1}{x {}^{3} } + 9 = 27 \\ \\ \\ x {}^{3} + \frac{1}{x {}^{3} } = 27 - 9 \\ \\ \\ x {}^{3} + \frac{1}{x {}^{3} } = 18 \\ \\ \\ Therefore \: \: \: \: \: \: \: \: \: x {}^{3} + \frac{1}{x {}^{3} } = 18 \\ \\ \\ 03 \\ \\ find \: \: \: \: \: \: x {}^{4} + \frac{1}{x {}^{4} } \\ \\ Answer \: \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 7 \\ \\ squaring \: \: on \: \: both \: \: sides \: \: we \: \: have \\ \\ (x {}^{2} + \frac{1}{x {}^{2} } ) = (7) {}^{2} \\ \\ \\ x {}^{4} + \frac{1}{x {}^{4} } + 2x {}^{2} \frac{1}{x {}^{2} } = 49 \\ \\ \\ x {}^{4} + \frac{1}{x {}^{4} } + 2 = 49 \\ \\ \\ x {}^{4} + \frac{1}{x {}^{4} } = 49 - 2 \\ \\ \\ x {}^{4} + \frac{1}{x {}^{4} } = 47 \\ \\ \\ Therefore \: \: \: \: \: \: x {}^{4} + \frac{1}{x {}^{4} } = 47$